4.1: Cosets and Lagranges Theorem
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Let \(G\) be a group. Let \(H \le G\) and \(g \in G\). Then
a) left coset of \(H\) with representative \(g\in G\) is denoted by \(gH=\{gh|h\in H\}\).
b) right coset of \(H\) with representative \(g\in G\) is denoted by \(Hg=\{hg|h\in H\}\).
Note: if \(G\) is abelian then\(H\) will be abelian, therefore \(gH=Hg, \; \forall g\in G\). The right cosets and left cosets are the same. Also, since \(H\) is a group, \(hH=H=Hh, \forall h \in H\).
Let \(G=\mathbb{Z}_6=\{0, 1, 2, 3, 4, 5\}\) with the operation addition (mod 6).
Let \(H=\{0,3\}\) which is a subgroup of \(\mathbb{Z}_6\). We must assure that \(H\) is, in fact a subgroup by ensuring there is closure, an identity exists, each element has inverse and operation on the set is associative. In this case clearly yes.
Left cosets of \(H\) will be:
0 (from G) + H = {0, 3}. Note: this is equal to H and also equal to 3 + H. Note: the count of
cosets are based on unique sets, thus 0+H=3+H counts as 1 left coset.
1 + H = {1, 4}. Note this is equal to 4 + H.
2 + H = {2, 5}. Note that this is also equal to 5+ H.
Therefore, there are 3 left cosets in \(H\).
Note, in this case, the right cosets will be the same sets as the left cosets. Thus \(\mathbb{Z}_6\) is commutative.
Let \(G=S_3=\{e, (1,2),(1,3),(2,3),(1,2,3),(3,2,1)\}\) with the operation composition.
Let \(H=A_3=\{e, (1,2,3),(3,2,1)\}\).
Without showing the work, \(eA_3=A_3\) and \(eA_3=(1,2,3)A_3=(3,2,1)A_3=A_3\).
Consider \((1,2)A_3=\{(1,2)e, (1,2)(1,2,3),(1,2)(3,2,1)\}\)
\(=\{(1,2), (2,3), (1,3)\}\).
\((1,3)A_3=\{(1,3)e, (1,3)(1,2,3),(1,3)(3,2,1)\}\)
\(=\{(1,3), (1,2), (2,3)\}\).
\((2,3)A_3=\{(2,3)e, (2,3)(1,2,3),(2,3)(3,2,1)\}\)
\(=\{(2,3), (1,3), (1,2)\}\).
Therefore \((1,2)A_3=(1,3)A_3=(2,3)A_3\).
Thus we have 2 left coset of \(H\), \(A_3\) and \(\{(2,3), (1,3), (1,2)\}\).
To calculate the right cosets, we have \(A_3G\). Applying the process similar to that to determine the left cosets, we obtain the right cosets which happen to be equal to the left cosets.
Let \(G\) be a group and let \(H\le G\). For \(g_1, g_2 \in G\), the following statements are equivalent:
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\(g_1H=g_2H\). Does this mean \(g_1=g_2\) or just that there is a g1 and g2 that will generate H?
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\(g_2^{-1}g_1H=H\).
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\(Hg_1^{-1}=Hg_2^{-1}\).
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\(g_2 \in g_1H\).
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\(g_1H\subseteq g_2H\).
Let \(G\) be a group and let \(H \le G\). Define a relation on \(G\) by \(g_1 \sim g_2\) is the same as \(g_1H=g_2H\). Then we can show that \(\sim\) is an equivalence relation (ie, it is reflexive, symmetric, and transitive). Then \(G= \underset{g \in G}{\cup} gH\) means either \(g_1H=g_2H\) or \(g_1H \cap g_2H= \{\}\).
Let \(G\) be a group and \(H \le G\).
The index of \(H\) in \(G=[G:H]\) Note: \([G:H]\) is an integer.
\(=\) the number of left cosets of H in G.
\(=\) the number of right cosets of H in G.
Let \(G\) be a finite group and \(H \le G\).
Then \(|G|=[G:H]|H|\). Note: The order of \(H\) divides the order of \(G\) and \([G:H]=\frac{|G|}{|H|}\).
Since \(|S_3|=6\) and let \(|K|=2\) then \([S_3:K]=\frac{6}{2}=3\), which is what we obtained in the previous example.
The converse of Lagrange's Theorem is not true.
Consider \(A_4\). \(6\) divides \(|A_4|\), but \(A_4\) has no subgroup of order \(6\).
\(A_4=\{e, (1\, 2\, 3), (1\, 3\, 2), (1\, 2\, 4), (1\, 4\, 2), (1\, 3\, 4), (1\, 4\, 3), (2\, 3\, 4), (2\, 4\, 3), (1\, 2)(3\, 4), (1\, 3)(2\, 4), (1\, 4)(2\, 3)\}\).
- Answer
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Assume that \(H\) is a subgroup of \(A_4\) such that \(|H|=6.\) Since \(|H|=6,\) at least one of the \(3-\) cycle in \(A_4\) must be in \(H\).Without loss of generality, we may assume \((1\, 2\, 3) \in H\). Since \(H\) is a subgroup of \(A_4\), h must contain \((1\, 2\, 3)^{-1}=(3\, 2\, 1)\).
Since \([A_4,H]=\frac{12}{6}=2\), \(gH=Hg, \; \forall g\in A_4\). Thus \(gHg^{-1}\in H, \; \forall g\in A_4.\)
Choose \(g=( 1\, 2\, 4) in A_4\). Then \((1\, 2\, 4)(1\, 2\, 3)(1\, 2\, 4)^{-1} =(1\, 2\, 4)(1\, 2\, 3)(4\, 2\, 1)=(2\, 4\, 3)\in H\). Since \(H\) is a subgroup of \(A_4\) , \((2\, 4\, 3)^{-1}=(3\, 4\, 2)\in H\). Now, \((2\, 4\, 3)(1\, 2\, 3)(2\, 4\, 3)^{-1} =(2\, 4\, 3)(1\, 2\, 3)(3\, 4\, 2)=(1\, 4\, 2)\in H.\)
Thus \(H=\{e, (1\, 2\, 3), (1\, 3\, 2), (1\, 2\, 4), (1\, 4\, 2), (1\, 3\, 4), (1\, 4\, 3), (2\, 3\, 4), (2\, 4\, 3), \cdots\}\). Thus \(H\) has more than \(6\) elements. This is a contradiction. Hence \(A_4\) has no subgroup of order \(6\).
Every group of prime order is cyclic.
If \(K \le H \le G\) then \([G:K]=[G:H][H:K]\).
From this theorem, we can prove that \(A_4\) has no subgroup of order 8. \(|A_4|=12\). From Lagrange’s theorem, the possible subgroups of \(A_4\) will have orders of 1, 2, 3, 4, 6, or 12. Thus there will not be a subgroup of order 8.