Prove that the center of a group is a normal subgroup [closed]

Solution 1:

As you said in a comment you already showed that it is normal. So I will only show that it is a subgroup.

Clearly it contains $e$, since $eg = ge$.

Now, we will show that it is closed. Let $a,b \in H$, we know that $\forall g: ag = ga$ and $gb = gb$. Thus, $gab = agb = abg$ and thus $ab \in H$.

Now we only have to show that every $h \in H$ has an inverse and we are done. Let $h \in H$, we know that $\forall g \in G: gh = hg$, thus $$\begin{align*}h^{-1}(gh)h^{-1} &= h^{-1}(hg)h^{-1}\\ h^{-1}g(hh^{-1}) &= (h^{-1}h)gh^{-1}\\ h^{-1}(ge) &= (eg)h^{-1}\\ h^{-1}g &= gh^{-1} \end{align*}$$

Which implies that $h^{-1} \in H$.

Solution 2:

I have two different solutions. These are probably too fancy for this problem, but they might be interesting.

For the first solution, define the map $f: G \rightarrow \operatorname{Aut}(G)$ by $g \mapsto \phi_g$, where $\phi_g(x) = gxg^{-1}$ for all $x \in G$. The map $f$ is a homomorphism and $\operatorname{Ker}(f) = Z(G)$. Thus $Z(G)$ is a normal subgroup since the kernel of a homomorphism is always a normal subgroup. The image $\operatorname{Im}(\phi)$ is called the inner automorphism group of $G$ and is denoted $\operatorname{Inn}(G)$.

For the second solution, recall that for a subgroup $H \leq G$ the normal core of $H$ in $G$ is defined as

$$\operatorname{core}_G(H) = \bigcap_{g \in G} gHg^{-1}$$

The subgroup $\operatorname{core}_G(H)$ is always a normal subgroup. This can be seen directly or by noticing that it is the kernel of the coset action induced by $H$. For any conjugacy class $C$ of $G$, let $t_c \in C$. Let $\mathscr{C}$ be the family of all conjugacy classes of $G$. Then \begin{align*} Z(G) &= \bigcap_{g \in G} C_G(g) \\ &= \bigcap_{C \in \mathscr{C}} \bigcap_{t \in C} C_G(t_c) \\ &= \bigcap_{C \in \mathscr{C}} \bigcap_{g \in G} C_G(gt_cg^{-1}) \\ &= \bigcap_{C \in \mathscr{C}} \bigcap_{g \in G} gC_G(t_c)g^{-1} \\ &= \bigcap_{C \in \mathscr{C}} \operatorname{core}_G(C_G(t_c)) \\ \end{align*}

Since $Z(G)$ is the intersection of normal subgroups, it is a normal subgroup.

Solution 3:

sxd has shown that it is a subgroup.

That it is normal follows from here:

Let $x\in Z(G)$ (center of $G$).

Then for any $g\in G$, $gxg^{-1}=gg^{-1}x=x\in Z(G)$.

This proves $Z(G)$ is a normal subgroup.