# Why do all elements commute in centre of an arbitrary finite group?

The section below is from wikipedia:

The center of G is always a subgroup of G. In particular:

1. $$Z(G)$$ contains the identity element of $$G$$, $$e$$, because, for all $$g \in G$$, it commutes with $$g$$, $$eg = g = ge$$, by definition;

2. If $$x$$ and $$y$$ are in $$Z(G)$$, then so is $$xy$$, by associativity: $$(xy)g = x(yg) = x(gy) = (xg)y = (gx)y = g(xy)$$ for each $$g \in G$$; i.e., $$Z(G)$$ is closed;

3. If $$x$$ is in $$Z(G)$$, then so is $$x^{−1}$$ as, for all $$g$$ in $$G$$, $$x^{−1}$$ commutes with $$g$$: $$(gx = xg) \implies (x^{−1}gxx^{−1} = x^{−1}xgx^{−1}) \implies (x^{−1}g = gx^{−1})$$. Furthermore, the center of $$G$$ is always a normal subgroup of $$G$$, as it is closed under conjugation, as all elements commute.

I do not understand the last bit, why do all elements commute in $$Z(G)$$?

My attemt was to try to get $$ab=ba$$: $$a,b \in Z(G) \implies \exists g,ga=ag, gb=bg\implies gagb=agbg$$, but this doesnt't seem to lead anywhere useful.

• Apply $g=b$..... – Randall Sep 21 '18 at 15:46
• Your "$\exists$" should be a "$\forall$." – Noah Schweber Sep 21 '18 at 15:50

The elements in the centre of a group, by definition, commute with all elements of the group $$G$$. Therefore they certainly commute with eachother.

Specifically, $$Z(G) := \{ z \in G | zg = gz \; \forall g \in G\}$$. If you have $$a,b \in Z(G)$$, then putting $$a$$ and $$b$$ into the definition above gives you what you want, right?

• In my notes I have this: centre $Z$ is defined to be the set of elements which each form their own conjugacy class. From here, is it obvious that they commute with each other? I think what this means that there is at least one element in $G$ with which they commute. Why can we say that "they commute with all", not "at least one"? – zabop Sep 21 '18 at 15:50
• Looking at your edit... I dont get why is it $Z(G) := \{ z \in G | zg = gz \; \forall g \in G\}$ and not $Z(G) := \{ z \in G | zg = gz \; \exists g \in G\}$ – zabop Sep 21 '18 at 15:52
• If an element $x \in G$ forms its own conjugacy class, then you have that $\{gxg^{-1} | g \in G\} = \{x\}$. In particular, $gxg^{-1} = x \; \forall g \in G$. Thus, if $x \in Z(G)$, $gx = xg$ for all $g \in G$. My alternative definition then drops out, and things follow. Your assertion that "there is at least one element in $G$ with which they commute" is not quite correct, although it is understandable why you're confused. I think the definition which I give above of the centre of a group is a little easier for less experienced people to understand. – Matt Sep 21 '18 at 15:54
• Additionally: It is worth you knowing that not only does your definition imply mine, but mine implies yours. They're equivalent. I've given you one implication, and you should see if you can get the reverse implication too. – Matt Sep 21 '18 at 15:57

Refer to @Matt's answer. The key step helping me to understand both the Q and that answer:

Two group elements $$g_1, g_2 \in G$$ are conjugate to each other ($$g_1 \sim g_2$$) if there exists any group element $$g \in G$$ s.t.

$$g_2 = gg_1g^{−1}$$.

Now, if $$g_1$$ is its own conjugacy class, then $$gg_1g^{−1}$$ has always has to give $$g_1$$ for any $$g$$ because if it does not, then $$g_1$$ would be in the same conjugacy class as the result of $$gg_1g^{−1}$$, so for any $$g$$, we have $$gg_1=g_1g$$, if $$g_1$$ forms its own conjugacy class.