Group of order 1183 is abelian if and only if contains an element of order 91 Let $G$ be a group such that $|G|=1183=7\cdot 13^2$. Show that $G$ is abelian if and only if $G$ has an element of order $91=7\cdot 13$.
What i did:
$7||G|\Rightarrow \exists x\in G : |x|=7$ and $13||G|\Rightarrow \exists y\in G : |y|=13$. Then $G$ abelian $\Rightarrow xy=yx$, So  $xy\in G:|xy|=7\cdot 13=91$.
Let $\exists x\in G : |x|=91$, consider the subgroup $H=<x>\leq G$. Also $13||G|\Rightarrow \exists y\in G : |y|=13$, so consider the subgroup $K=<y>\leq G$.
Now what i wanted to show $G=HK$, but i am not sure whether $H\cap K=1$ and either anyone of $H,K$ is normal.
Is my approach correct? If not what are the shortcomings? If anyone could point out my mistakes and hint me a correct method then i'd appreciate his/her help. Thank you.
 A: You're on the right track with the first part (but, as suggested in the comments, you should be careful with your justifications). For the second part, I'm not sure your approach will be fruitful: even if you show $HK$ is a subgroup equal to $G$, it's not clear to me that this is useful. 
To help you out, I'll sketch a solution to a similar problem. You can work out the details of my argument, and adapt the strategy to your problem.
Problem: Let $G$ be a group of order $3179=11\cdot 17^2$. If $G$ contains an element of order $11\cdot 17 =187$, then $G$ is abelain.
Let $x$ be an element of order $187$. I claim that $x$ commutes with all elements of some Sylow $17$-subgroup of $G$. Given this, it's easy to show that $x$ lies in the center $Z(G)$ of $G$. This implies that $G/Z(G)$ has size $1$ or $17$, and so $G$ is abelian.
To prove the claim, note that the Sylow theorems imply $G$ has a unique subgroup $T$ of order $17^2$. This allows us to consider the action of $\langle x^{17} \rangle \simeq \mathbb Z/ 11 \mathbb Z $ on  $T$ by conjugation. Nontrivial orbits of this action have size $11$.
The order $17$ subgroup $\langle x^{11} \rangle$ lies in $T$, and its elements are fixed points of the action. If the action is non-trivial, it has no other fixed points and $T - \langle x^{11} \rangle$ is a disjoint union of nontrivial orbits. But $|T - \langle x^{11} \rangle| = 16\cdot 17$, which is not divisible by $11$. Hence the action is trivial, and all elements of $T$ commute with both $x^{11}$ and $x^{17}$. The claim follows.
