I am tasked with the following problem:
Prove that any subgroup of index 2 is normal (invariant)
This is the proof that I have been given:
Let’s say that group $G = \{E, g_2, . . . g_{|G|}\}$ has a subgroup $H = \{E, g_2, g_3, . . . , g_{|G|/2}\}$ of index 2. $H$ has half as many elements as $G$ has. Since the distinct cosets of $H$ do not share any elements, and each coset has $|H|$ elements, we can conclude that there exists only $|G|/|H| = 2$ distinct left cosets, and for same reasons, only 2 distinct right cosets of $H$. Now that we know $H$ only has two distinct right and left cosets, $\color{red}{\text{we must show that the left cosets are the same as the right cosets}}$.
The subgroup itself is always a coset as we can multiply the identity with the subgroup. That is, the left and right cosets $HE = EH = H$. Since the cosets do not share any elements, the other left and right coset is the set containing all elements of $G$ not in $H$, let’s call it $H' = \{g_{|G|/2+1}, . . . , g_{|G|−1}, g_{|G|}\}$. Therefore the two left cosets are $H$ and $H'$, and the two right cosets are also $H$ and $H'$. The left and right cosets of $H$ are the same, and so $H$ is an invariant subgroup.
More generally, for $h \in H$, $hH = Hh = H$ and for $g' \in H'$, $g'H = Hg' = H'$ . So in any case, $gH = Hg$ for any $g$ and the subgroup is invariant.
End of proof.
I'm okay right up until the point I marked in red. Why do we need to show that the left cosets and right cosets are the same?
Remark:
I don't see this as a 'homework style' question since I am asking for an explanation on a particular proof, and there is a tag for that. But, if you see fit to retag it with homework-and-exercises then fine by me.
Best Answer
The left cosets of $G$ with respect to $H$ are the subsets of $G$ of the form $gH$, with $g\in G$, whereas the right cosets are those of the from $Hg$. Now, suppose that $H$ is a normal subgroup of $G$. This means that, for each $g\in G$, $g^{-1}Hg\subset H$. Actually, since $G$ is finite, it means that $g^{-1}Hg=H$. This, in turn, is equivalent to $Hg=gH$. So, if $H$ is a normal subgroup of $G$, $\{Hg\mid g\in G\}=\{gH\mid g\in G\}$; in other words, the set of right cosets is equal to the set of left cosets. It's not hard to prove that if the set of right cosets is equal to the set of left cosets, then $H$ is a normal subgroup of $G$.