If $H \leq G$ and $H \subset Z(G)$, the center of $G$, is $H \trianglelefteq G$? This is probably a very dumb question. Is it true that, in general, if $H$ is a subgroup of a group $G$, and $H \subset Z(G)$, the center of $G$, does it follow that $H$ is normal in $G$?
What I know so far that could potentially be useful:


*

*$Z(G)$ is normal in $G$.

*$H$ is normal in $G \Leftrightarrow gHg^{-1} \subseteq H$ for all $g \in G$.


This is part of an intermediate step for a homework problem. 
 A: This can be understood intuitively with group actions. Say $G$ acts on a set $X$:


*

*A subset $Y\subseteq X$ is pointwise fixed if $gy=y$ for all $g\in G$.

*A subset $Y\subseteq X$ is setwise fixed if $gY:=\{gy:y\in Y\}=Y$ for all $g\in G$.


The group $G$ acts on itself by conjugation. Then:


*

*A subset $H\subseteq G$ is central if $~[G,H]=1 ~~\Leftrightarrow~H\subseteq Z(G)~ \Leftrightarrow~  H$ is pointwise fixed.

*A subset $H\subseteq G$ is normal if it is setwise fixed.
For general group actions, pointwise fixed is much stronger than ($\Rightarrow$) setwise fixed. Therefore we may directly conclude that central ($H\subseteq Z(G)$) implies normal ($H\trianglelefteq G$) in this case.
A: Yes. As a hint for how to prove it, let $h \in H$ and use the fact that $h \in Z(G)$ to simplify $ghg^{-1}$.
A: Since any $h\in H\le Z(G)$ then $hx=xh$ for all $x\in G$, so $xhx^{-1}=h\in H$ i.e. 
$$xHx^{-1}\subseteq H,$$
which is enough to warranty that $H$ is normal in $G$.  
A: Here is how I think about it. If you are in the center, you commute with everything. If $H$ is normal, you have to be able to conjugate an element $h \in H$ by $g \in G$ and land back in $H$, in other words $ghg{-1} \in H$. But $h$ is in the center, so you have $ghg^{-1}=gg^{-1}h=h$ which is certainly in $H$. So $H$ is normal.
