Let $G$ be a group, $H\neq G$ and $H\le G$ and $a\in G$. which ones are always correct?
a) $|a|=|a^{-1}|$
b) H and G has the same unit
c) $|H|\neq0$
d) if $a^{12}=a^2$ then |a| is even number
e) $\langle \langle a\rangle\cup H\rangle$ is a subgroup of G.
b and c is correct because $H\le G$
for d, if $|G|=m, m|(12-2)$ and $|a|$ divides $m$ so |a| can be 5
a)
$e=(aa^{-1})^n=a^na^{-n}$, so if one is $e$ the other must also be $e$, so it is true.
b)
the identity is the only idempotent element in $G$, (since $xx=x\implies x^{-1}xx=x^{-1}x\implies x=e$) and the unit of $H$ must be idempotent also.
c)
Since $H$ has an identity it has at least one element.
d)
$e$ satisfies $|e^{12}|=|e^2|$ and $|e|=1$. So no.
For another example consider $\mathbb Z_{31}$ and $a=2$. Notice $2(2)=4$ and $12(2)=24$ both have order $31$, while $2$ also has order $31$.
e)
By definition $\langle S \rangle $ is a subgroup of $G$ for any subset $S$ of $G$. the subset $\langle a\rangle \cup H$ is no exception.