If $H$ is a normal subgroup of group $G$ of odd order and $|H|=5$ . show that $H \subseteq Z(G)$
Attempt: If $|H|=5$ and if $H$ is normal, then $H$ must be a normal cyclic subgroup.
$\implies H = \{e,a,a^2,a^3,a^4\}$ for some $a \in G$
Since, $H$ is normal, $\implies gH = Hg \implies g a^i = a^j g$
And now, we need to prove that $i=j$.
$G$ is of odd order and $|H|=5 \implies$ index of $H$ in $G = [G:H] = |G|/|H|=2m-1$ is odd
$\implies g^{2m-1} \in H~~\forall~ g \in H$
How do I proceed further? Help will be appreciated.
Thank you
Attempt 2: By the $N/C$ Theorem : Since $H$ is a normal subgroup of $G \implies N(H) =G.$
Now, $H$ is cyclic of order $5 \implies H \subseteq C(H)$ and $|Aut(H)|=1,2$ or $4$
CASE $1$ : $|Aut(H)|=1 \implies |G|=|C(H)| \implies C(H)=G$
CASE $2$: $|Aut(H)|=2 \implies C(H)$ is normal in $G ~~ i.e.~~ C(H) \vartriangleleft G$.
CASE $3$: $|Aut(H)|=4 \implies G/C(H) \approx Z_2 \bigoplus Z_2 $ or $~ G/C(H) \approx Z_4$
We need to show that $\forall h \in H,~~ hg~=~gh ~~\forall g \in G$. I am kind of stuck again. Help will be appreciated. Thank you