Let G be an abelian group and $a,b\in G$ be two distinct elements with a and b or order $2$.
Show that $H=\{e,a,b,ab\}$ forms a subgroup and write out its multiplication table.
Justify why all the elements in $H$ must be distinct.
Prove that $$K=\{g \in G : |g| \leq 2\}.$$ Show that this can fail if $G$ is not abelian.
I can do part 1 (show $H$ has associativity, inverses, closure, and the identity element), but for parts 2 and 3 how do I show distinctness, assume two are equal and make a contradiction? Because I can't see how that will get me somewhere. And for 3, where do I begin?
$G$ have $4$ elements, one of which must be the unit $e$ and there must be another element $e\neq a\in G$.
Now, there is another element in the group $b$ with $b\neq a$ and $b\neq e$ .
Finally, consider the element $ab\in G$: $$ ab=a\implies a^{-1}ab=a^{-1}a\implies b=e $$
which is false. $$ ab=b\implies abb^{-1}=bb^{-1}\implies a=e $$
which is also false.
and $$ ab=e\implies b=a^{-1} $$
but $a$ is of order $2$ so $a=a^{-1}$ and thus $b=a$ which is again false.
We conclude $ab$ is the fourth element in $G$.
If I understood what part $3$ is asking you: Consider the Dihedral group of a square $G=D_{8}$,
We have it that $D_{8}=\langle s,rs\rangle$ since $r=rs\cdot s\in\langle s,rs\rangle$ but although $s,rs$ are both of order $2$ $$ G\neq K=\{g\in G\mid|g|\leq2\} $$
since $r\in D_{8}$ is of order $4$.