Here we say that a group $G$ is periodic if every element of $G$ has finite order. $G$ needs not be abelian.

Problem. Let $G$ be a group with center $Z<G$. Prove that if every element in $G\setminus Z$ has finite order, then $G$ is periodic.

My proof is as follows: Assume $z\in Z$ has infinite order.

Select an element $a\in G\setminus Z$. By construction, there exists some $b\in G\setminus Z$ such that $ab\neq ba$.

On the one hand, the element $za$ is not in the center as $$(za)b=z(ab)\neq z(ba)=(zb)a=(bz)a=b(za).$$

On the other hand, the element $za$ also has infinite order: Suppose $(za)^m=(za)^n$. By assumption, $a$ has finite order, say $k$, then we have $$(za)^{mk}=(za)^{nk}\implies z^{mk}a^{mk}=z^{nk}a^{nk}\implies z^{mk}=z^{nk},$$ contradicting the assumption that $z$ has infinite order.

However, by assumption, since $za\in G\setminus Z$, it should have finite order, giving a contradiction.

Therefore, every element in the center $Z$ also has finite order and hence $G$ is periodic. $\square$

Hope anyone can help to check my proof. Different approaches are highly welcomed.

  • $\begingroup$ What if $G$ is abelian though? Then $Z=G$ and you can't proceed. For instance, if $G=\Bbb Z$. $\endgroup$
    – user403337
    Jan 10, 2021 at 0:26
  • $\begingroup$ Perhaps I misunderstood. You meant to assume $G$ isn't abelian. Got it. $\endgroup$
    – user403337
    Jan 10, 2021 at 0:39
  • $\begingroup$ @ChrisCuster: I think that the use of $\lt$ is meant to indicate a proper subgroup (as opposed to $\leq$). $\endgroup$ Jan 10, 2021 at 2:12

1 Answer 1


The argument is fine, but perhaps overly complicated; and you don’t actually need to cast it as a proof by contradiction: you can do a direct proof. Note that you never use the assumption that $z$ has infinite order until you are done proving that it has finite order: that’s typical of what I call “fake proofs by contradiction” (fake because they aren’t really proofs by contradiction, they are direct proofs or proofs by contrapositive that are dressed up to appear to be proofs by contradiction).

So, let me point some places where it can be improved:

  1. The argument about $za$ can be done simpler: in general, if $H$ is a subgroup of $G$, $h\in H$, and $g\in G$ with $g\notin H$, then $gh,hg\notin H$. So if $z\in Z$ and $a\in G\setminus Z$, then $za\notin Z$.

  2. Once you know that, we prove that $z$ has finite order directly, essentially how you do it: we know $a$ and $za$ both have finite order, say $n\gt 1$ and $m\gt 1$. Then, since $za=az$, we have: $$1 = ((za)^m)^n = (za)^{nm} = z^{nm}a^{nm} = z^{nm}(a^n)^m = z^{nm}1^m = z^{nm}.$$ Therefore, $z$ has finite order dividing $nm$.


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