0
$\begingroup$

Prove that if $H$ is a subgroup of index $2$ in a finite group $G$, then $gH = Hg \; \forall \; g \in G$.

I know that $H$ itself is one coset of the subgroup and the other is the compliment of the subgroup, but I don't really understand why the second coset is the compliment. I know that the union of the cosets must be $G$, but how do we know that we can't say, for example, $2H \cup H \equiv G$? Why do we know for sure that $H'\cup H$ is $G$?

I also know that the number of left and right cosets are identical.

$\endgroup$
2
$\begingroup$

Remember, Langrange's theorem? There was an equivalence relation $a \sim b$ if $a^{-1}b \in H$ So $G\backslash H$ are the equivalence classes induced by $a \sim b$ if $a^{-1}b \in H$ In other words, the cosets represent the different equivalence classes induced by this equivalence relation. What do we know from equivalence classes? They $partition$ $G$, so if $aH$ and $bH$ are two different cosets then $aH \cap bH = \emptyset$ Therfore, if H has index 2 that means that $G\backslash H$ has two cosets. One coset you know is H and the other coset must have $a \in G$ such that $a \notin H$ which is just $H'$ These must also be the right cosets, hence $gH =Hg $ $\forall g \in G$

$\endgroup$
0
$\begingroup$

As you already know the number of left cosets is equal to the number of right ones, just take into account that

$$aH=H\iff a\in H$$

$$Ha=H\iff a\in H$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.