I have the following question, where the topic being tested is cosets, order and Lagrange's theorem:
Suppose that every element $x$ in a group $G$ satisfies $x^2 = e$. Prove that $G $is abelian.
Show also that if $H$ is any subgroup of $G$ and $g ∈ G\backslash$ then $K = H ∪ gH$ is a subgroup of $G$.
Show further that $K$ is isomorphic to $H × C_2$.
Deduce that if G is finite then G is isomorphic to $(Z_2)^n$ for some non-negative integer n.
I've done the first part fairly easily, and the second as well just by a quick use of the subspace test. However I'm really stumped on how to go about the third part.
For the fourth part, I get the feeling it will have to be some kind of induction, since I know that I could extend the result from the second part to $K$ to make another new subgroup, which will then be isomorphic to $C2 \times C2 \times H$, and so on. Since $Z_2^n$ and $C_2$ are isomorphic, its just a notational difference.
Any help would be much appreciated!
Best Answer
The following map suggests itself as candidate for an isomorphism $\phi\colon K\to H\times C_2$: $$ \phi(x)=\begin{cases}(x,0)&\text{if $x\in H$}\\(gx,1)&\text{if $x\in gH$}\end{cases}$$ Check it.
For part 4, note that $G$ contains some subgroups of the form $C_2^k$ (for example the trivial subgroup, where $k=0$). Among those subgroups, let $H\approx C_2^n$ be one of maximal size. If there exists $g\notin H$ we see that $K:=H\cup gH\approx H\times C_2\approx C_2^n\times C_s\approx C_2^{n+1}$ is a strictly larger such group (this is where we use finiteness!), contradicting maximality of $H$. Hence no $g\notin H$ exists, i.e. $G=H\approx C_2^n$. (This use of a maximal subgroup of the given type is in fact nothing else but a convoluted way of doing induction on the order of $G$, so your idea was the right one)