# H is a subgroup of index 2 in finite group G, then every left coset of H is a right coset as well

I know this means that there are two cosets. I also know that one must be H itself. This means that the remaining coset(s) must be equal for right and left. Also since there is only one possible coset, this means that for all elements, a,b in G, then $aH = bH$ and $Ha = Hb$, since each element in G acting on the subgroup must produce the same set. This means $aH = Ha$ I believe. From there $H = aHa^{-1}$, which given the information I am not sure if this good or how to use it. Any help would good.

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looks suspiciously like a homework problem. – user17762 Nov 11 '10 at 22:27
Not really. aH = bH iff a and b are in the same coset. Also, what is actually your question? Because it looks like you already know that the implication in the title is correct. – Marek Nov 11 '10 at 22:36
It was a homework problem. Am I not allowed to ask them? I knew the implication was correct, however I am not very good at proof and this is new material. If it helps I did spend quite a while on it. – Metahominid Nov 12 '10 at 10:41
For what it's worth, this has a generalization to any prime p: if p is the smallest prime factor of #G, then any subgroup of index p in G has every left coset equal to a right coset. The proof for general p is different from the argument given here when p = 2. – KCd Nov 12 '10 at 10:57

The (left) cosets of $H$ partition the group $G$; since there are two cosets and one of them is $H$, that means that, whatever the other one is (call it $\mathcal{C}$), as a set, you have that $H\cap\mathcal{C}=\emptyset$ and $H\cup \mathcal{C}=G$. These two together tell you that the other one must be equal, as a set, to the complement of $H$ inside of $G$; that is, $\mathcal{C}=G-H$. So the two cosets are $H$ and $G-H$. Of course, $G-H$ can be written as $xH$ for some $x$; in fact, for any $x\notin H$ you have $xH = G-H$.

But the exact same argument holds for right cosets: one of them is $H$, and the other one must be, as a set, the complement of $H$. So the two right cosets are $H$ and $G-H$, and $G-H$ can be written as $Hy$ for some $y$; in fact, for any $y\notin H$ you have $Hy=G-H$.

Now note that in both cases, the cosets are: 1. $H$, the coset of the elements of $H$; and 2. $G-H$, the coset of the elements not in $H$.

So each of the two left cosets is also a right coset and vice-versa.

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Thank you both very much – Metahominid Nov 12 '10 at 2:02

You say "Also since there is only one possible coset, this means that for all elements, a,b in G, then aH = bH and Ha = Hb, since each element in G acting on the subgroup must produce the same set." but this is not true. Let a be the identity and b not be in H.

But earlier you say "I know this means that there are two cosets. I also know that one must be H itself." And if all cosets are the same size, what can you say about the one that isn't H?

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If $|G:H| = 2$ and $1 \neq g \in G$, then $$G = H \; \dot{\cup} \; gH \;\;\; and \;\;\; G = H \; \dot{\cup} \; Hg$$ Therefore $$gH = Hg \;\;\;\; \forall 1 \neq g \in G$$ Of course $$gH = Hg \;\;\;\; for \;\; 1 = g$$ Therefore $$gH = Hg \;\;\;\; \forall g \in G$$

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According to my textbook, you're correct. if N is a subgroup of group G, then these are equivalent: 1. N is normal
2. left cosets of N=right cosets of N
3. For all a in G, aN=Na

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If you haven't covered normal subgroups yet, just ignore the last sentence 'By an .... H is normal."

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