Let $g \in N_G(H)$ an element of order $5$. Compute the order of $H \langle g \rangle$.

Question

Let $G = A_5$ and $H=\langle(1,2,3,4,5)\rangle$. Let $g \in N_G(H)$ an element of order $5$. Compute the order of $H \langle g \rangle$.

I think I can use the second isomorphism theorem to solve this problem. I find that $|HK| = \frac {|H||\langle g \rangle|}{|H \cap \langle g \rangle|}$, but I don't know how to find $H \cap \langle g \rangle$.

Is anyone is able to give me a hint?

Answer 1

Both of the subgroups $H$ and $<g>$ are of order 5 which is prime, so either they are equal, or their intersection is trivial.

EDIT: full solution

Since $g$ has order 5 and it is in $A_n$ it must be a 5-cycle. If $g\notin H$, then $H\cap <g>=\{e\}$ since they are both subgroups of prime order, and then you get that $<H,g>$ is a group of order 25 (because $<H,g>\cong H\rtimes <g>$ using the normality of $H$ in $<H,g>$). This is not possible, since $|A_5|=5!/2=60$ is not divisible by 25. It therefore follows that $g$ must be in $H$.