# Let $G$ be a group and $H = \{ g \in G \mid gx = xg \ \forall x \in G \}$. Show that $H \leqslant G$ and that $H$ is commutative.

Let $$G$$ be a group and $$H = \{ g \in G \mid gx = xg \ \forall x \in G \}$$. Show that $$H \leqslant G$$ and that $$H$$ is commutative.

How can I show that $$H \leqslant G$$? This seems to me that I would want to show that $$H$$ is a subgroup of $$G$$? That would seem to imply the statement $$H \leqslant G$$. So I would need to show that $$H$$ is non empty and closed under products and inverses?

Let $$h\in H$$, then $$hh^{-1} = e_G$$ right? So the inverse is there. How can I show the closure under products here?

Recall that for $$H$$ to be a subgroup of $$G$$, it suffices to prove that for all $$a, b \in H$$, we have $$ab^{-1} \in H$$.

Suppose $$a, b \in H$$. We have $$x^{-1} b = b x^{-1}$$ for all $$x \in G$$, which is equivalent to $$b^{-1}x = xb^{-1}$$. We now multiply with $$a$$ from the left and get $$ab^{-1}x=axb^{-1}$$. Now use that $$a$$ commutes and associativity, and arrive at

$$(ab^{-1})x = x(ab^{-1}),$$

which finishes the proof.

EDIT: In order to use this subgroup test, it is also necessary to show that $$H$$ is non-empty. This is trivial since the identity element of $$G$$ will always be in $$H$$.

• It does not suffice to show that $a,b\in H\implies ab^{-1}\in H$; if $H$ is empty to begin with it certainly never will be a subgroup. The precise statement of the one-step subgroup test is that if $H$ is non-empty and $a,b\in H\implies ab^{-1}\in H$ then $H$ is a subgroup. Feb 2, 2021 at 13:43
• That is of course correct, @mrtaurho. I have now added an edit pointing this out.
– Möb
Feb 2, 2021 at 13:47
• Aren't we supposed to show that $a$ commutes and not use it before that?
– user869998
Feb 2, 2021 at 13:54
• The so-called one-step subgroup test is as mrtaurho has described above. So we need to assert that $H$ is non-empty and show that $a, b \in H \implies ab^{-1} \in H$. To do this you assume that $a, b \in H$. Then you are allowed to use that these two elements commute.
– Möb
Feb 2, 2021 at 14:11

To check closure under inverses you have to show that $$h^{-1}\in H$$. You only know that $$h$$ is invertible as element of $$G$$, that is there is $$h^{-1}\in G$$ such that $$hh^{-1}=e_G$$ but it might be the case that $$h^{-1}\in G\setminus H$$, i.e. that the inverse is not in the subset $$H\subseteq G$$. Similarily, you have to show that $$e_G\in H$$ and that if $$h_1,h_2\in H$$ implies that $$h_1h_2\in H$$.

Let's take a closer look at the inverses. We are given $$h\in H$$ and want to show that $$h^{-1}\in H$$ which is to say that $$h^{-1}x=xh^{-1}$$ for all $$x\in G$$. We are given that $$xh=hx$$ since by assumption $$h\in H$$. Now, first multiply from the left by $$h^{-1}$$ and then from the right by $$h^{-1}$$ to get:

$$xh=hx\quad\implies\quad h^{-1}xh=x\quad\implies\quad h^{-1}x=xh^{-1}$$

Now, show that $$xe_G=e_Gx$$ for all $$x\in G$$ and that if $$h_1,h_2\in H$$ then $$x(h_1h_2)=(h_1h_2)x$$ for all $$x\in G$$ to complete the proof.

Aside: the subgroup you are dealing with is called the center of $$G$$, denoted by $$Z(G)$$ and measures "how abelian" the group is. We have, for instances $$G=Z(G)$$ if and only if $$G$$ is abelian.

If $$gx=xg$$ for all $$x\in G$$ i.e. $$g\in H$$ and $$hx=xh$$ for all $$x\in G$$ i.e. $$h\in H$$ then $$(gh)x=g(hx)=g(xh)=(xh)g=x(hg)=x(gh)$$for all $$x\in G$$ and thus $$gh\in H$$ i.e. $$H$$ is closed under multiplication.

$$H$$ is the $$\it{center}$$ of $$G$$ and is denoted $$\mathcal Z(G)$$. Let $$g$$ be variable over $$G$$ and note $$xgx^{-1}=g\iff x^{-1}gx=g$$ \begin{align}&\therefore\;x,y\in H\implies g=xgx^{-1}=y^{-1}gy\implies (xy^{-1})g(yx^{-1})=x(y^{-1}gy)x^{-1}=g\\&\therefore\;x,y\in H\implies xy^{-1}\in H\\ &\therefore H\leq G\end{align}