Cosets and Lagrange’s Theorem And Examples Coset
Definition. Let (H,.) be a subgroup of the group (G,.).
Let a ∈ G. Then the set aH = {ah | h ∈ H} is called a left coset of H in G generated by a and the set Ha = {ha | h ∈ H} is called a right coset of H in G generated by a.
Here the operation in G is denoted multiplicatively. Also aH, and Ha are called cosets of H generated by a in G.
Since every element of aH or Ha is in G, aH and Ha are complexes of G.
If e is the identity in G, then eH = {eh \ h ∈ H} = {h | h e H} = H and He = {he | he H} = H. Hence the subgroup of G is itself a left and a right coset of H in G.
If e is the identity in G, it is also the identity in H.
Therefore, for a ∈ G,e ∈ H we have ea ∈ Ha and ae ∈ aH.
Hence the left coset or the right coset of H generated by a is non-empty. Further a ∈ Ha, a ∈ aH and \(\mathbf{H} a \cap a \mathbf{H} \neq \phi\).
If group G is abelian, then for every h ∈ H, we shall have ah = ha. Hence aH = Ha. However, even if G is not abelian, also we may have aH = Ha or aH ≠ Ha.
Note
1. The left or right coset of any subgroup in a group is called residue class modulo the subgroup of the group.
2. If the operation in G is denoted additively, then the left subset of H in G is generated by a, denoted by a + H = {a + h | h ∈ H}
i.e. a + H =.{a + h | h ∈ H}.
Similarly, the right coset of H is G generated by a.
= H + a = {h + a | h ∈ H}.
3. Let H be a subgroup of the group G and a, b ∈ G.
Then (1) a (bH) = (ah) H and (Hb)a = H(ba).
(2) x ∈ aH => yx ∈ y(aH) for y ∈ G => yx e (ya) H.
4. The clement a is called the coset representative of aH (Ha).
| aH |, | Ha | denotes the number of elements in aH, Ha respectively,
Example 1. Consider the group of symmetries of the square
i.e. \(\left(\mathbf{D}_4, o\right)\) where \(\mathbf{D}_4=\left\{r_{90}, r_{180}, r_{360}, x, y, d_1, d_2\right\}\)
(H,o) where \(\mathbf{H}=\left\{r_{180}, r_{360}, x, y\right\}\) is a subgroup of \(\left(\mathbf{D}_4, o\right)\).
Cosets Definition And Examples In Group Theory
Then all the left cosets of H in G are
\(r_{90} \mathbf{H}=\left\{r_{90} \circ r_{180}, r_{90} \circ r_{360}, r_{90} \circ x, r_{90} \circ y\right\}\) \(=\left\{r_{270}, r_{90}, d_2, d_1\right\}\)
\(r_{180} \mathbf{H}=\left\{r_{180} o r_{180}, r_{180} o r_{360}, r_{180} o x, r_{180} o y\right\}\) \(=\left\{r_{360}, r_{180}, x, y\right\}=\mathbf{H}\)
\(r_{270} \mathbf{H}=\left\{r_{270} o r_{180}, \ldots \ldots\right\}\) \(=\left\{r_{90} ; r_{270}, d_1, d_2\right\}\)
\(r_{360} \mathbf{H}=\left\{r_{360} o r_{180}, \ldots \ldots\right\}\) \(=\left\{r_{180}, r_{360}, x, y\right\}=\mathbf{H}\)
\(x \mathbf{H}=\left\{x \circ r_{180}, \ldots \ldots\right\}\) \(=\left\{x, y, r_{360}, r_{180}\right\}=\mathbf{H}\)
\(y \mathbf{H}=\left\{y o r_{180}, \ldots \ldots\right\}\) \(=\left\{x, y, r_{180}, r_{360}\right\}=\mathbf{H}\)
\(d_1 \mathbf{H}=\left\{d_1 o r_{180}, \ldots \ldots\right\}\) \(=\left\{d_2, d_1, r_{270}, r_{90}\right\}\)
\(d_2 \mathbf{H}=\left\{d_2 o r_{180}, \ldots \ldots\right\}\) \(=\left\{d_1, d_2, r_{90}, r_{270}\right\}\)
We have two distinct left cosets, namely,
∴ \(\left\{r_{90}, r_{270}, d_1, d_2\right\} \text { and }\left\{r_{180}, r_{360}, x, y\right\}=\mathbf{H}\)
These two may be taken as \(r_{90} \mathbf{H} \text { and } \mathbf{H}\)
Obviously \(r_{90} \mathbf{H} \cap \mathbf{H}=\phi \text { and } r_{90} \mathbf{H} \cup \mathbf{H}=\mathbf{D}_4\)
Similarly, we can have all the right cosets of \(\mathrm{H} \text { in } \mathbf{D}_4\).
Note. If a ∈ H, then aH = H = Ha.
Lagrange’s Theorem In Group Theory Explained With Examples
Example 2. Let G be the additive group of integers.
Now G = {…… -3, -2, -1, 0, 2, 3,…….. } and 0 is the identity in G. Also G is abelian.
Let H be a subset of G where elements of H are obtained by multiplying each element of G by 3 (say) i.e., H = { ……..,-9, -6,-3, 0,-3, 6, 9……}
Clearly, H is a subgroup of (G,+). (∵ \(n_1, n_2 \in \mathbf{H} \Rightarrow n_1-n_2 \in \mathbf{H}\))
Since G is abelian, the left coset of H of an element in G = right coset of H in G.
∴ 0 + H = H = {………, -9,-6,-3, 0, 3, 6,……}.
Since 1 ∈ G, 1 + H = { ………..,-8, -5, -2, 1, 4, 7,……}
Since 2 ∈ G, 2 + H = {………..,-7, -4, -1, 2, 5, 8,……}
Observe that (1) 3 + H = 6 + H = …… = 0 + H ,
4 + H = 7 + H = ……… = 1 + H,
5 + H = 8 + H = …….. = 2 + H.
(2) 0 + H, 1 + H, 2 + H are disjoint.
(3) 0 + H ∪ 1 + H ∪ 2 + H = G
Cosets and Lagrange’s Theorem And Examples Properties of Cosets
Theorem 1. H is any subgroup of a group (G,) and h ∈ G. Then h ∈ H iff hH = H = Hh.
Proof. (1) h ∈ H to prove that hH = H = Hh.
Let ‘h’ be an arbitrary element of H. Then is an arbitrary element of hH.
Since H is a subgroup of G, h, h’ ∈ H => hh’∈ H
Thus every element of hH is also an element of H.
∴ hH ⊆ H …………………(1)
Again \(h^{\prime}=e h^{\prime}=\left(h h^{-1}\right) h^{\prime}=h\left(h^{-1} h^{\prime}\right) \in h \mathbf{H}\)
(∴ e is the identity in H, \(h \in \mathbf{H} \Rightarrow h^{-1} \in \mathbf{H} \text { and } h^{-1} \in \mathbf{H}, h^{\prime} \in \mathbf{H} \Rightarrow h^{-1} h^{\prime} \in \mathbf{H}\))
∴ H ⊆ hH
From (1) and (2), hH = H.
Similarly, we can prove H = Hh
∴ h ∈ H
∴ hH = H = Hh.
(2) Let hH = H = Hh . To prove that he H.
Now h ∈ G. Since h = he, h ∈ hH.
But hH = H
∴ h ∈ H
Similarly Hh = H => h ∈ H
∴ hH = H= Hh => h ∈ H
Step-By-Step Guide To Understanding Cosets And Lagrange’s Theorem
Theorem 2. If a,b are any two elements of a group (G,.) and H any subgroup of G, then \(\mathbf{H} a=\mathbf{H} b \Leftrightarrow ab^{-1} \in \mathbf{H}\) and aH = bH ⇔ \(a^{-1} b \in \mathbf{H}\).
Proof. \(a \in \mathbf{H} a, \mathbf{H} a=\mathbf{H} b \Rightarrow a \in \mathbf{H} b \Rightarrow a b^{-1} \in(\mathbf{H} b) b^{-1}\)
\(\Rightarrow a b^{-1} \in \mathbf{H}\left(b b^{-1}\right) \Rightarrow a b^{-1} \in \mathbf{H} e \Rightarrow a b^{-1} \in \mathbf{H}\)Now \(a b^{-1} \in \mathbf{H} a b^{-1}=\mathbf{H} \Rightarrow \mathbf{H} a b^{-1} b=\mathbf{H} b\)
\(\Rightarrow \mathbf{H} a\left(b^{-1} b\right)=\mathbf{H} b \Rightarrow \mathbf{H} a e=\mathbf{H} b \quad \Rightarrow \mathbf{H} a=\mathbf{H} b\)Similarly we can Prove that \(a \mathbf{H}=b \mathbf{H} \Leftrightarrow a^{-1} b \in \mathbf{H}\)
Note. \(\text { If } a b^{-1} \in \mathbf{H} \text { then }\left(a b^{-1}\right)^{-1} \in \mathbf{H} \Rightarrow\left(b^{-1}\right)^{-1} a^{-1} \in \mathbf{H} \Rightarrow b a^{-1} \in \mathbf{H}\)
Similarly \(a^{-1} b \in \mathbf{H} \Rightarrow b^{-1} a \in \mathbf{H}\)
Solved Examples Of Cosets In Mathematics
Theorem 3. If a,b are any two elements of a group G and H any subgroup of G, then \(a \in b \mathbf{H} \Leftrightarrow a \mathbf{H}=b \mathbf{H} \text { and } a \in \mathbf{H} b \Leftrightarrow \mathbf{H} a=\mathbf{H} b\)
Proof. \(a \in b \mathbf{H} \Rightarrow b^{-1} a \in b^{-1} b \mathbf{H}\)
\(\Rightarrow b^{-1} a \in e \mathbf{H} \Rightarrow b^{-1} a \in \mathbf{H}\) \(\Rightarrow b^{-1} a \mathbf{H}=\mathbf{H} \Rightarrow b b^{-1} a \mathbf{H}=b \mathbf{H} \Rightarrow a \mathbf{H}=b \mathbf{H}\)Converse: Let aH = bH
∴ \(a \in a \mathbf{H} \Rightarrow a \in b \mathbf{H}\)
Similarly, other results can be proved.
Proof Of Lagrange’s Theorem With Examples
Theorem 4. Any two left (right) cosets of a subgroup are either disjoint or identical.
Proof. Let H be a subgroup of group G. Let aH and bH be two left cosets of H in G. If aH and bH are disjoint, there is nothing to prove.
If \(a \mathbf{H} \cap b \mathbf{H} \neq \phi\), then there exists at least one element c such that c ∈ aH and c ∈ bH.
Let \(c=a h_1 \text { and } c=b h_2 \text { where } h_1, h_2 \in \mathbf{H}\)
∴ \(a h_1=b h_2 \Rightarrow a h_1 h_1^{-1}=b h_2 h_1^{-1} \Rightarrow a e=b\left(h_2 h_1^{-1}\right) \Rightarrow a=b\left(h_2 h_1^{-1}\right)\)
Since H is a subgroup, \(h_2 h_1^{-1} \in \mathbf{H} \cdot \text { Let } h_3=h_2 h_1^{-1}\)
∴ \(h_3 \in \mathbf{H}\)
Now \(a=b h_3\)
∴ \(a \mathbf{H}=b h_3 \mathbf{H}=b \mathbf{H}\) (∵ \(h_3 \in \mathbf{H} \Rightarrow h_3 \mathbf{H}=\mathbf{H}\) )
Two left cosets are identical if they are not disjoint.
∴ \(a \mathbf{H} \cap b \mathbf{H}=\phi \quad \text { or } \quad a \mathbf{H}=b \mathbf{H}\)
Similarly, we can prove that \(\mathbf{H} a \cap \mathbf{H} b=\phi \text { or } \mathbf{H} a=\mathbf{H} b\).
Cor. H is any subgroup of a group G. If the cosets aH, bH, cH, … are all disjoint, then \(\mathbf{G}=\mathbf{H} \cup a \mathbf{H} \cup b \mathbf{H} \cup c \mathbf{H} \ldots\) where H is the cost corresponding to the identity element in G.
Also \(\mathbf{G}=\mathbf{H} \cup \mathbf{H} a \cup \mathbf{H} b \cup \mathbf{H} c \cup \ldots\)
Congruence Modulo H
Definition. Let (G,.) be a group and (H, .) be a subgroup of G. For a, b ∈ G, if \(b^{-1} a \in \mathbf{H}\) we say that a ≡ b (mod H)
Theorem 5. H is a subgroup of group G, for a,b ∈ G the relation a ≡ b (mod H) is an equivalence relation.
Proof. (1) Reflexive: Let e be the identity in (G,.).
Since H is a subgroup of G, e is the identity in H.
Let a ∈ G. Since \(a^{-1} a=e \text {, we have } a^{-1} a \in \mathbf{H}\)
∴ a ≡ a (mod H) => relation is reflexive.
(2) Symmetric: Let a ≡ b (mod H) for a, b ∈ G.
∴ \(b^{-1} a \in \mathbf{H} \Rightarrow\left(b^{-1} a\right)^{-1}, \in \mathbf{H} \Rightarrow a^{-1} b \in \mathbf{H}\)
b ≡ a (mod H )=> relation is symmetric.
(3) Transitive: Let a ≡ b (mod H) and b ≡ c (mod H) for a,b,c ∈ G.
∴ \(b^{-1} a \in \mathbf{H} \text { and } c^{-1} b \in \mathbf{H} \Rightarrow\left(c^{-1} b\right)\left(b^{-1} a\right) \in \mathbf{H}\)
\(\Rightarrow c^{-1}\left(b b^{-1}\right) a \in \mathbf{H} \Rightarrow c^{-1}(e a) \in \mathbf{H}\)\(\Rightarrow c^{-1} a \in \mathbf{H} \Rightarrow a \equiv c(\bmod \mathbf{H}) \Rightarrow\) relation is transitive.
Since the congruence modulo H is reflexive, symmetric, and transitive, it is an equivalence relation.
Note: Let H be a subgroup of group G and a ∈ G, then the equivalence class containing a w.r.t. the equivalence relation ( ≡ mod H ) is denoted by \(\bar{a}\).
Cosets And Lagrange’s Theorem In Abstract Algebra
Theorem 6. Let (H, .) be a subgroup of a group (G,.). For a ∈ G, let the equivalence class \(\bar{a}=\{x \in \mathbf{G} / x≡a(\bmod \mathbf{H})\}\). Then \(\bar{a}=a \mathbf{H}\)
Proof. To prove that \(\bar{a}=a \mathbf{H}\)
Let e be the identity in G.
∴ e is also the identity in H.
\(x \in \bar{a} \Leftrightarrow x \equiv a(\bmod \mathbf{H})\) \(\Leftrightarrow a^{-1} x \in \mathbf{H}\)\(\Leftrightarrow a^{-1} x=h \in \mathbf{H} \text { for some } h \in \mathbf{H}\)\(\Leftrightarrow a\left(a^{-1} x\right)=a h \in a \mathbf{H} \text { for some } h \in \mathbf{H}\)
\(\Leftrightarrow\left(a a^{-1}\right) x=a h \in a \mathbf{H} \text { for some } h \in \mathbf{H}\) \(\Leftrightarrow e x=a h \in a \mathbf{H} \text { for some } h \in \mathbf{H}\) \(\Leftrightarrow x=a h \in a \mathbf{H} \text { for some } h \in \mathbf{H}\) \(\Leftrightarrow x \in a \mathbf{H}\)∴ \(\bar{a}=a \mathbf{H}\)
Note 1. The equivalence relation a ≡ b (mod H) induces a partition in G which is nothing but the left coset decomposition of G. w.r.t. H. No left coset of H in G will be empty. Every element of G belongs to one and only one left coset of G.
2. The relation in G, defined by a ≡ b (mod H) if \(a b^{-1} \in \mathbf{H}\), is an equivalence relation. This relation induces a partition in G which is nothing but the right coset decomposition of G.
Theorem 7. Let (H,) be a subgroup of a group (G.). Then there exists a bijection between any two left cosets of H in G.
Proof. Let aH,bH be two left cosets of H for a, b ∈ G.
Define \(f: a \mathbf{H} \rightarrow b \mathbf{H} \text { such that } f(a h)=b h \text { for } h \in \mathbf{H}\)
For \(h_1, h_2 \in \mathbf{H}, a h_1, a h_2 \in a \mathbf{H} \text { and } b h_1, b h_2 \in b \mathbf{H}\)
Now \(f\left(a h_1\right)=f\left(a h_2\right) \Rightarrow b h_1=b h_2 \Rightarrow h_1=h_2 \Rightarrow a h_1=a h_2\)
∴ \(f \text { is } 1-1\)
Now \(b h \in b \mathbf{H} \Rightarrow \exists h \in \mathbf{H} \text { such that } b h \in b \mathbf{H}\)
⇒ \(\exists h \in \mathbf{H} \text { such that } a h \in a \mathbf{H}\)
∴ For ah ∈ aH , f(ah) = bh
∴ f is onto.
∴ f is a bijection and there exists 1 — 1 correspondence between any two left cosets of H in G.
Note 1. Let H be a subgroup of a finite group G. Since there is 1—1 correspondence between any two left cosets of H, every left coset has the same number of elements including H (v H is also a left coset).
2. The above theorem can be proved between two right cosets. Also, every right coset of H of a finite group G has the same number of elements including H. (∵ H is also a right coset).
Properties Of Cosets Explained With Solved Problems
Theorem 8. If w is a subgroup of a group G then there is a one-to-one correspondence between the set of all distinct left cosets of H in G and the set of all distinct right cosets of H in G.
Proof. In G, let \(\mathrm{G}_1\) = the set of all distinct left cosets
and \(\mathrm{G}_2\) = the set of all distinct right cosets.
Define a mapping \(f: \mathbf{G}_1 \rightarrow \mathbf{G}_2 \text { such that } f(a \mathbf{H})=\mathbf{H} a^{-1} \forall a \in \mathbf{G}\)
For: \(\text { Let } a \mathbf{H}, b \mathbf{H} \in \mathbf{G}_1\)
Now \(a \mathbf{H}=b \mathbf{H} \Rightarrow b^{-1} a \in \mathbf{H} \Rightarrow\left(b^{-1} a\right)^{-1} \in \mathbf{H}\)
⇒ \(a^{-1}\left(b^{-1}\right)^{-1} \in \mathbf{H} \quad \Rightarrow \mathbf{H} a^{-1}=\mathbf{H} b^{-1}\)
⇒ \(f(a \mathbf{H})=f(b \mathbf{H})\)
⇒ \(f \text { is one-one : Let } a \mathbf{H}, b \mathbf{H} \in \mathbf{G}_1\)
∴ \(f(a \mathbf{H})=f(b \mathbf{H}) \Rightarrow \mathbf{H} a^{-1}=\mathbf{H} b^{-1}\)
⇒ \(a^{-1}\left(b^{-1}\right)^{-1} \in \mathbf{H} \Rightarrow a^{-1} b \in \mathbf{H} \Rightarrow\left(a^{-1}b\right)^{-1} \in \mathbf{H}\)
⇒ \(b^{-1} a \in \mathbf{H} \Rightarrow a \mathbf{H}=b \mathbf{H}\)
∴ \(f \text { is } 1-1\).
f is onto Let \(\mathbf{H} a \in \mathbf{G}_2\).
Since \(a \in \mathbf{G}, a^{-1} \in \mathbf{G}\)
∴ \(a^{-1} \mathbf{H} \in \mathbf{G}_1 \text { and } f\left(a^{-1} \mathbf{H}\right)=\mathbf{H}\left(a^{-1}\right)^{-1}=\mathbf{H} a\)
∴ \(f \text { is onto }\)
There is a one-to-one correspondence between \(\mathbf{G}_1 \text { and } \mathbf{G}_2\).
Note 1. If H is a subgroup of a finite group G, then the number of distinct left cosets of H in G is the same as the number of distinct right cosets of H in G.
2. Since H is common to both the set of left cosets of H of a finite group G and the set of right cosets of H of the finite group G, the number of elements in a left coset of H is equal to the number of elements in a right coset of H.
Cosets and Lagrange’s Theorem And Examples Index of a subgroup of a finite group.
Definition. If H is a subgroup of a finite group G, then the number of distinct left (right) cosets of H in G is called the index of H in G. It is denoted by (G: H) or \(\)
Lagrange’s Theorem
Theorem .9. The order of a subgroup of a finite group divides the order of the group.
Proof. Since H is a subgroup of a finite group G, H is finite.
(1) If H = G, then O(H) / O(G) .
(2) If H ≠ G, let O(G) = n and O(H) = m
We know that every right coset of H in G has the same number of elements and the number of right cosets of H in G is finite.
Also since H = He, H is a right coset of H in G.
∴ If Ha, Hb, Hc, are right cosets of H in G, then
O (Ha) = O(Hb) = O(Hc) = …….. = O (H) = m
Let the number of distinct right cosets of H of G be k
All these right cosets are disjoint and induce a partition of G.
∴ \(\mathbf{O}(\mathbf{G})=\mathbf{O}(\mathbf{H} a)+\mathbf{O}(\mathbf{H} b)+\mathbf{O}(\mathbf{H} c)+\ldots \ldots+\mathbf{O}(\mathbf{H})(k \text { terms })\)
= \(m+m+m+\ldots+m(k \text { times }) \Rightarrow n=k m \Rightarrow k=\frac{n}{m}\)
∴ \(\mathbf{O}(\mathbf{H}) \text { divides } \mathbf{O}(\mathbf{G}) \text { i.e. } \mathbf{O}(\mathbf{H}) / \mathbf{O}(\mathbf{G})\)
Note 1. Lagrange’s theorem can also be proved by taking the right cosets of H in G.
2. Lagrange’s theorem deals with finite groups only.
Let O(G) = n. If m is not a divisor of n, then there can be no subgroup of G of order m.
3. Since \(k=\frac{n}{m}\) number of the distinct left (right) cosets of H in \(\mathbf{G}=\frac{|\mathbf{G}|}{|\mathbf{H}|}\).
= \(\frac{\text { order of the group } \mathbf{G}}{\text { order of the subgroup } \mathbf{H} \text { of } \mathbf{G}}=\text { Index of } \mathbf{H} \text { in } \mathbf{G}=(\mathbf{G}: \mathbf{H})\)
4. The converse of Lagrange’s theorem is not true.
(1) Consider G = {1, – 1,i, – i} . Clearly. G is a group of order 4 w.r.t. multiplication. Since 2 is a divisor of 4 i.e. the order of the group G, let us examine whether a complex H (of order 2) of G, which is a subgroup of G, exists.
Consider a complex \(\mathbf{H}_1=\{i, \quad-i\}\)
Since – i. i = 1 and since \(1 \notin \mathbf{H}_1 . \mathbf{H}_1\) is not a subgroup of G.
Again consider a complex \(\mathbf{H}_2=\{1,-1\}\). Clearly, \(\mathbf{H}_2\) is a subgroup of G.
∴ In conclusion, even if m is a divisor of n, a subgroup of order m in G need not exist.
(2) G is a finite group of order 6. Since 3 is a divisor of 6 i.e. the order of the group G, let us examine whether a complex H (of order 3) of G, which is a subgroup of G, exists.
Consider a complex \(\mathbf{H}_1=\left\{r_0, f_1, f_2\right\} \text { of } \mathbf{G}\)
Since \(f_1 \circ f_2=r_1\) and since \(r_1 \notin \mathbf{H}_1, \mathbf{H}_1\) is not a subgroup of G.
Again consider a complex \(\mathbf{H}_2=\left\{r_0, r_1, r_2\right\}\).
Clearly \(\mathbf{H}_2\) is a subgroup of G with identity \(r_0\) and with \(r_0^{-1}=r_0, r_1^{-1}=r_2\) and
∴ \(r_2^{-1}=r_1\).
∴ In conclusion, even if m is a divisor of n, a subgroup of order m in G need not exist.
Thus the converse of Lagrange’s Theorem does not hold.
Cor. : The order of an element of a finite group divides the order of the group.
Theorem 10. Suppose H and K are subgroups of a group G such that K ≤ H ≤ G and suppose (H: K) and (G: H) are both finite. Then (G: K) is finite, and (G: K) = (G:H)(H: K)
Proof: H and K are Subgroups of a group G such that K ≤ H ≤ G and suppose (H: K) and (G: H) are both finite.
(G: H) = the index of subgroup H in G is the number of distinct left cosets of H in G and (H: K) = the index of subgroup K in H is the number of distinct left cosets of K in H.
Thus by Lagrange’s Theorem : \((\mathbf{G}: \mathbf{H})=\frac{|\mathbf{G}|}{|\mathbf{H}|} \text { and }(\mathbf{H}: \mathbf{K})=\frac{|\mathbf{H}|}{|\mathbf{K}|}\)
∴ \((\mathbf{G}: \mathbf{H})(\mathbf{H}: \mathbf{K})=\frac{|\mathbf{G}|}{|\mathbf{H}|} \cdot \frac{|\mathbf{H}|}{|\mathbf{K}|}=\frac{|\mathbf{G}|}{|\mathbf{K}|}=(\mathbf{G}: \mathbf{K})\) implying that (G: K) is finite and (G: K) = (G: H) (H: K)
OR :
Suppose that the collection of distinct left cosets of H in \(\mathbf{G}=\left\{a_i \mathbf{H}: i=1,2, \ldots \ldots, r\right\}\) and the collection of distinct left cosets of K in \(\mathbf{H}=\left\{b_j \mathbf{K}: j=1,2, \ldots \ldots, s\right\}\).
Now we show that \(\left\{a_i b_j \mathbf{K}: i=1 ; 2, \ldots . . r, j=1,2, \ldots . s\right\}\) is the collection of distinct left cosets of K in G.
G \(=\bigcup_{i=1 \text { to r }} a_i \mathbf{H}, a_i \in \mathbf{G} \text { and } \mathbf{H}=\bigcup_{i=1 \text { to s }} b_j \mathbf{K}, b_j \in \mathbf{G}\)
Now \(x \in \mathbf{G} \Rightarrow x \in \bigcup_i a_i \mathbf{H} \Rightarrow x=a_1 h, h \in \mathbf{H} \text { and }\)
h\(\in \mathbf{H} \Rightarrow h \in \bigcup_j b_j \mathbf{K} \Rightarrow h=b_j \mathbf{K}, k \in \mathbf{K}\)
∴ \(x=a_i h=a_i b_j \mathbf{K}, k \in \mathbf{K} \Rightarrow x \in \bigcup_{i, j} a_i b_j \mathbf{K} \Rightarrow \mathbf{G}=\bigcup_{i, j} a_i b_j \mathbf{K}\)
Now we show \(a_i b_j \mathbf{K}=a_{i^{\prime}} b_{j^{\prime}} \mathbf{K} \Leftrightarrow i=i^{\prime}, j=j^{\prime}\)
If \(i=i^{\prime}, j=j^{\prime}, \text { then } a_i b_j \mathbf{K}=a_{i^{\prime}} b_{j^{\prime}} \mathbf{K}\)
If possible \(xa_i b_j \mathbf{K}=a_{i^{\prime}} b_{j^{\prime}} \mathbf{K} \text { when } i=i^{\prime}, j \neq j^{\prime}\)
Then \(b_j \mathbf{K} \cap b_{j^{\prime}} \mathbf{K}=\phi \Rightarrow a_i b_j \mathbf{K} \cap a_{i^{\prime}} b_{j^{\prime}} \mathbf{K}=\phi\)
=> it is a contradiction.
∴ \(j=j^{\prime}\)
If possible\(a_i b_j \mathbf{K}=a_{i^{\prime}} b_{j^{\prime}} \mathbf{K} \text { when } i \neq i^{\prime}, j=j^{\prime}\)
Then \(b_j \mathbf{K}=b_{j^{\prime}} \mathbf{K} \text { and } a_i \mathbf{H} \cap a_{i^{\prime}} \mathbf{H}=\phi \Rightarrow a_i b_j \mathbf{K} \cap a_{i^{\prime}} b_{j^{\prime}} \mathbf{K}=\phi\)
=> it is a contradiction.
∴ \(i=i^{\prime}\)
When \(i \neq i^{\prime}, j=j^{\prime}, a_i \mathbf{H} \cap a_{i^{\prime}} \mathbf{H}=\phi \text { and }\)
∴ \(b_i \mathbf{K} \cap b_i, \mathbf{K}=\phi \Rightarrow a_i b_j \mathbf{K} \cap a_{i^{\prime}} b_{j^{\prime}} \mathbf{K}=\phi\)
∴ \(a_i b_j \mathbf{K}=a_{i^{\prime}} b_{j^{\prime}} \mathbf{K} \Leftrightarrow i=i^{\prime}, j=j^{\prime}\)
Thus G is the collection of distinct left cosets of K in G.
Hence (G: K) is finite and (G: K) = (G: H) (H: K)
(\(\text { Euler’s } \phi \text {-function }\) It is the function \(\phi: Z^{+} \rightarrow Z^{+}\) defined as (1) \(\text { For } 1 \in Z^{+}, \phi(1)=1\) and (2) for \(n(>1) \in Z^{+}, \phi(n)=\) the number of positive integers less than n and relatively prime to n.)
Theorem 11. If n is a positive integer and a is an integer relatively prime to n then \(a^{(x)}≡1(\bmod n) \text { where } \phi \text { is the Euler’s } \phi \text {-function. }\)
Proof: Let x be any integer. Let [x] denote the residue class of the set of integers mod n.
G = {[a]/a is an integer relatively prime to n}.
Then G is a group of order ∅(n) with respect to the multiplication of residue classes. The identity in G is [1].
⇒ \([a] \in \mathbf{G} \Rightarrow[a]^{0(\mathbf{G})}=[1] \Rightarrow[a]^{\phi(n)}=[1]\)
a a a a } . . . to \(\phi((n) \text { times }]\) =[1]
⇒ \(\left[a^{\phi(n)}\right]=[1] \Rightarrow a^{\phi(n)} \equiv 1(\bmod n)\)
This theorem is known as Euler’s theorem.
Normalizer Of An Element Of A Group
Definition. If a is an element of a group G, then the normalizer on a in G is the set of all those elements of G which commute with a. The normalizer of an in G is denoted, by N (a) where N (a) = {x ∈ G / ax = xa).
The normalizer N (a) is a subgroup of G.
Note. If e is the identity in group G, \(e x=x e=x \forall x \in \mathbf{G} \Rightarrow \mathbf{N}(e)=\mathbf{G}\)
Cosets and Lagrange’s Theorem And Examples Solved Problems
Example. 1. Use Lagrange’s Theorem to prove that a finite group cannot be expressed as the union of two of its proper subgroups.
Solution: Let G be a finite group of order n. Assume that \(\mathbf{H} \cup \mathbf{K}=\mathbf{G}\) where H, K are two proper subgroups of G.
Since e ∈ H and e ∈ K at least one of H, K (say H) must contain more than half the number of elements of G.
Let O(H) = p
∴ \(\frac{n}{2}<p<n\) (∵ H is a proper subgroup of G )
∴ n is not divisible by p which contradicts Lagrange’s theorem.
Hence our assumption that \(\mathbf{H} \cup \mathbf{K}=\mathbf{G}\) is wrong.
∴ A finite group cannot be expressed as the union of two of its proper subgroups.
Example. 2. Show that two right, cosets Ha, Hb of a group G only if the two left cosets \(a^{-1} \mathbf{H}, b^{-1} \mathbf{H}\) of G are distinct.
Solution: Suppose that (Ha) = (Hb).
H\(a=\mathbf{H} b \Leftrightarrow a b^{-1} \in \mathbf{H} \Leftrightarrow a b^{-1} \mathbf{H}=\mathbf{H}\)
⇒ \(\Leftrightarrow a^{-1} a b^{-1} \cdot \mathbf{H}=a^{-1} \mathbf{H}\)\(\Leftrightarrow b^{-1} \mathbf{H}=a^{-1} \mathbf{H} \Leftrightarrow a^{-1} \mathbf{H}=b^{-1} \mathbf{H}\)
∴ Ha, Hb are distinct iff \(a^{-1} \mathbf{H} \text { and } b^{-1} \mathbf{H}\) are distinct.
Example. 3. Show that every finite group of prime order does not have any proper subgroup.
Solution: Let G be a finite group of order n where n is prime.
If possible, let H be a subgroup of order m, say
Then m ≤ n. But by Lagrange’s theorem m is a divisor of n.
Also since n is prime, either m = 1 or m = n.
∴ H = {e} or H = G. But these two are improper subgroups of G.
∴ Any group of prime order does not have any proper subgroup.
Note. Thus the total number of subgroups of a group of prime order is 2.
Example. 4. \(\mathbf{P}_3=\left\{f_1, f_2, f_3, f_4, f_5, f_6\right\}\) is a non-abelian group over S.
⇒ \(\mathbf{H}=\left\{f_1, f_2\right\}\) is a subgroup of \(\mathbf{P}_3\).
Let us form the left cosets of H in \(\mathbf{P}_3\).
⇒ \(f_1 \mathbf{H}=\mathbf{H}, f_2 \mathbf{H}=\mathbf{H}, f_3 \mathbf{H}=\mathbf{H}=\left\{f_3, f_6\right\}, f_4 \mathbf{H}=\left\{f_4, f_5\right\}\)
⇒ \(f_5 \mathbf{H}=\mathbf{H}=\left\{f_5, f_4\right\}, f_6 \mathbf{H}=\left\{f_6, f_3\right\}\)
Thus we get only three distinct left cosets.i.e. H, \(f_3 \mathbf{H}, f_4 \mathbf{H} \text { of } \mathbf{H} \text { in } \mathbf{P}_3\)
Thus \(\mathbf{P}_3=\mathbf{H} \cup f_3 \mathbf{H} \cup f_4 \mathbf{H}\) and index of subgroup H in \(\mathbf{P}_{\mathbf{3}} \text { is } 3\).
Observe that the number of elements in each left coset is the same as in H.
Further \(f_3 \mathbf{H} \neq \mathbf{H} f_3 \text { since } \mathbf{H} f_3=\left\{f_3, f_5\right\}\).
We can observe similar results by taking all the right cosets of H in G.
Note. \(\mathbf{A}_3=\left\{f_1, f_5, f_6\right\}\) is a commutative subgroup of \(\mathbf{P}_3\).
Two distinct left cosets of \(\mathbf{A}_3\) are \(\mathbf{A}_3, f_2 \mathbf{A}_3\) where \(f_2 \mathbf{A}_3=\left\{f_2, f_3, f_4\right\}\).
Also \(\mathbf{P}_2=\mathbf{A}_3 \cup f_2 \mathbf{A}_3\) and index of a subgroup of \(\mathbf{A}_3 \text { in } \mathbf{P}_3 \text { is } 2\).
Example. 5. Let H be a subgroup of group G and let \(\mathbf{T}=\{x \in \mathbf{G} / x \mathbf{H}=\mathbf{H} x\}\). Show that T is a subgroup of G.
Solution: H is a subgroup of group G.
Let \(x_1, x_2 \in \mathbf{T}\)
∴ \(x_1 \mathbf{H}=\mathbf{H} x_1, x_2 \mathbf{H}=\mathbf{H} x_2\)
Now \(x_2 \mathbf{H}=\mathbf{H} x_2 \Rightarrow x_2^{-1}\left(x_2 \mathbf{H}\right) x_2^{-1}=x_2^{-1}\left(\mathbf{H} x_2\right) x_2^{-1}\)
⇒ \(\mathbf{H} x_2^{-1}=x_2^{-1} \mathbf{H} \Rightarrow x_2^{-1} \in \mathbf{T}\)
Also \(\left(x_1 x_2^{-1}\right) \mathbf{H}=x_1\left(x_2^{-1} \mathbf{H}\right)=x_1\left(\mathbf{H} x_2^{-1}\right)=\left(x_1 \mathbf{H}\right) x_2^{-1}\)
= \(\left(\mathbf{H} x_1\right) x_2^{-1}=\mathbf{H}\left(x_1 x_2^{-1}\right) \Rightarrow x_1 x_2^{-1} \in \mathbf{T}\)
Thus \(x_1, x_2 \in \mathbf{T} \Rightarrow x_1, x_2^{-1} \in \mathbf{T}\)
∴ T is a subgroup of G.
Conjugate Element Of A Group Self
Definition. (G,.) is a group and a ∈ G such that \(a=x^{-1} a x \forall x \in \mathbf{G}\). Then a is called a self-conjugate element of G. A self-conjugate element is sometimes called an invariant element.
Here \(a=x^{-1} a x \Rightarrow x a=a x \forall x \in \mathbf{G}\)
Cosets and Lagrange’s Theorem And Examples The centre of a group.
Definition. The set Z of all self-conjugated elements of group G is called the centre of group G.
Thus \(\mathbf{Z}=\{z \in \mathbf{G} / z x=x z \forall x \in \mathbf{G}\}\)
If G is abelian, then the centre of G is G. (Vide Theorem 18 Chapter 5)
Cosets and Lagrange’s Theorem And Examples Exercise 4
1. If H = {1, – 1} and G = {1, – 1, i, – i] then prove that (H, .) is a subgroup of the group (G,.). Find all the right cosets of H in G.
Solution:
1 \(\mathbf{H}=\mathbf{H},(-1) \mathbf{H}=\{-1,1\}, i \mathbf{H}=\{i,-i\},(-i) \mathbf{H}=\{-i, i\}\)1
2. Prove that \(\) is a subgroup of \(\). Find the left cosets of the above subgroup in \(\). Find the index of the subgroup in G.
Solution:
Given
If H = {1, – 1} and G = {1, – 1, i, – i]
0 + \({ }_{15} \mathbf{H}=\{0,3,6,9,12\}=\mathbf{H}, 1+{ }_{15} \mathbf{H}=\{1,4,7,10,13\}, 2+_{15} \mathbf{H}=\{2,5,8,11,14\}\)
3+ \({ }_{15} \mathbf{H}=\mathbf{H}, 4+_{15} \mathbf{H}=1+_{15} \mathbf{H}=1+_{15} \mathbf{H}, 5+_{15} \mathbf{H}=2+_{15} \mathbf{H} \text {, etc. }\)
3. (1) Determine the coset decomposition of the additive group of integers relative to a subgroup of all integral multiples of 4 = 4Z.
(2) Find all co-sets and index of the subgroup < 4 > of \(\mathrm{z}_{12}\).
Solution:
(1) \(\mathbf{Z}=(0+\mathbf{H}) \cup(1+\mathbf{H}) \cup(2+\mathbf{H}) \cup(3+\mathbf{H})\)
(2) \(\mathbf{Z}_{12}=0+{ }_{12} \mathbf{H} \cup 1+{ }_{12} \mathbf{H} \cup 2+_{12} \mathbf{H} \cup 3+_{12} \mathbf{H}\) where H = {0, 4, 8} = <4>. The index of the subgroup H of\(\mathrm{Z}_{12}\) is 4.