Let G be a group such that for some integer $m>1$, $(ab)^m=a^mb^m, \; \forall a,b \in G$, and let $G^m=\lbrace a^m: a\in G\rbrace$. Show that $G^m \triangleleft G$ and order of each element of ${G}/{G^m}$ is finite.
$\underline{Claim: G^m \triangleleft G}$
Let $a$ be an arbitrary element of G and choose $h$ in $G^m$,$\;h$ has the form $h=b^m$ for some $b \in G$. Now, consider:
$$ab^ma^{-1}=\underbrace{(aba^{-1})(aba^{-1})(aba^{-1})...(aba^{-1})}_{m\;times}=(aba^{-1})^m\in G^m$$ It follows that for an arbitrary $a$ in $G$ we have $aG^ma^{-1} \subseteq G^m$. This implise $G^m \triangleleft G$.
$\underline{Claim:\text{Every element of } {G}/{G^m} \text{has finite order}}$
Since $G$ is a group and $G^m$ is a normal subgroup of $G$ we can talk about $ {G}/{G^m}$ as a quotient group(factor group). WLOG, let $g^m \in G^m$ then
$$g^mG^m = G^m \text{, since } aH=H \iff a\in H $$
This implies that $$g^m =e$$ Implies every element in a quotient group has finite order.
Do I make any mistake? Thanks in advance!