# Subset of elements of a given order in a group

I'm almost certain this is true (I have even given a proof) but I keep getting this strange feeling that something is not quite right so I will ask...

Let $G$ be an abelian group and let $r$ be a positive integer.

Then set of elements of order $r$ in $G$ together with the identity $e$ forms a subgroup of $G$.

I will be very grateful if someone can just confirm this for me.

Thanks!

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In general, if $r$ is not prime then the answer is no, because for any proper divisor $q | r$ you can find an element of order $q$ in the cyclic subgroup generated by an element of order $r$. If $r$ is prime then it is true.

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This is not true. The cyclic group of order $4$ along with the integer $r=4$ is a counterexample. There are plenty of other counterexamples too. To convince yourself that this fact is not true, try to find another counterxample.

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No. The set of elements of order $4$ (if such exist) together with $0$ do not form a group. Note that if $a$ is of order $4$, then $a+a$ is of order $2$. The fact that you had to add $0$ extra should have made you suspicious.

However, the set of elements $a$ such that $r\cdot a=0$, does form a subgroup: If $r\cdot a=0$ and $r\cdot b=0$ then $r\cdot(a+b)=0$ and $r\cdot (-a)=0$ (and of course $r\cdot 0=0$).

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I think the best you can say is that the set of elements of exponent $r$ is a subgroup.

This is easy to prove, because by definition this set is $\{ g \in G : g^r=1\} = \ker \phi$, where $\phi(x)=x^r$ is a homormorphism $G\to G$, since $G$ is abelian.

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