$G$ be a group that has finite number of not normal subgroups. Show that every infinite subgroup of $G$ is normal subgroup.

Let $G$ be a group that has finite number of not normal subgroups. Show that every infinite subgroup of $G$ is normal subgroup.

I would appreciate if someone can give me a hint!

My attempt: I think I solve it!

Assume $H$ be an infinite and not normal subgroup of $G$, so exist a member like $g$ that $gHg^{-1}\nsubseteq H$ , let consider $ghg^{-1}=x \in gHg^{-1}-H$ so $gxg^{-1}\notin \ \lt{x}\gt$ because $x=g^{-1}x^{n}g={(g^{-1}xg)}^{n}=h^n\in H$ and it is contradiction. so $\lt {x} \gt$ is not normal.

If $gHg^{-1}-H$ is infinite $\lt{x}\gt$ is not normal when $x\in gHg^{-1}-H$ and because $G$ has a finite number of not normal subgroups so exists distinct $x_1,x_2,\dots$ that $\lt{x_1}\gt=\lt{x_2}\gt=\lt{x_3}\gt=\dots$ so $\lt{x_1}\gt$ has infinite generator but every cyclic group has finite genrator and its contradiction.

So $gHg^{-1}-H$ is finite now let $\alpha\in gHg^{-1}-H$ and $gHg^{-1} \cap {H}=K$ so $$\alpha K \subseteq gHg^{-1}= (gHg^{-1}-H )\cup K$$

and since $gHg^{-1}-H$ is finite and $\alpha K$ is infinite so $\alpha K \cap K \neq \emptyset$ so implies $\alpha K=K$ and $\alpha \in H$ and its contradiction.

• Your set $K$ is by definition empty. – M.U. Aug 27 '16 at 15:06
• @M.U. you are right i edit it – user362389 Aug 27 '16 at 17:15