If $H \lhd G$ and $|H|=2$, show $H \subseteq Z(G)$.
Is this true when $|H|=3$?
I know that since $H \lhd G$ that means $\forall g \in G$ that $Hg=gH$ and $|H|=2$ that $H=\{1,h\}$ .
I think I need to show that $h \in Z(g)$ and then somehow look at cosets, I am just not sure how.
On
Let $h$ be the generator of $H$. Since $H$ is a normal subgroup of $G$, $g^{-1} h g$ must be in $H$ for every $g$ in $G$. Since $g^{-1} h g$ must still have order 2, it follows that $g^{-1} h g = h$ for every $g$ in $G$. But this means $h$ commutes with every element of $G$, so is in the center of $G$.
If $H$ has order 3, this argument fails. Conjugating a generator $h$ of $H$ by an arbitrary element of $G$, it's possible that we'll get back $h^{-1}$. Indeed, this can happen when $G$ is the nonabelian group of order 6, $S_3$. The unique subgroup of order 3 in $S_3$ is not central.
Let $g\in G$, $Ad_g:H\rightarrow H$ defined by $Ad_g(h)=ghg^{-1}$ is an automorphism, since $|H|=2$, $Ad_g$ is the identity. To see this write $H=\{1,h\}$ where $1$ is the neutral, $Ad_g(1)=1$, since $Ad_g$ is a bijection, $ad_g(h)=h$. and $gh=hg$.
$A_3$ has order 3 and is not in the center of $S_3$.