Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $G$ an abelian group of order $10$ contains an element of order $5$, how can I show that $G$ must be a cyclic group.

share|cite|improve this question
up vote 2 down vote accepted

By Cauchy's theorem we know that there exists an element y in G with order 2. Let x be an element of G of order 5.

Claim: $|xy|=10$


$(xy)^{|xy|}=e$, therefore $x^{|xy|}y^{|xy|}=e$ this implies: $$x^{2|xy|}y^{2|xy|}=e^2=e$$ $$x^{2|xy|}=e$$ Hence 5 divides 2|xy|, thus 2 divides |xy|. Similarly, 5 divides |xy|. Therefore, |xy|=10 (because |xy| divides 10)

share|cite|improve this answer
You can shorten the argument by saying that in an abelian group the order of $xy$ is $\frac{|x||y|}{\gcd(|x|,|y|)}$ – Jernej Nov 18 '12 at 10:25
Since the problem statement includes that an element of order 5 exists as premise (which would also follow by Cauchy), I suspect that Cauchy must not be used. – Hagen von Eitzen Nov 18 '12 at 10:41

The following does not make use of Cauchy's theorem.

$G$ contains exactly one element of order $1$, namely $e$. If $g_1,g_2\in G$ have order $5$, then either $\langle g_1\rangle=\langle g_2\rangle$ or $\langle g_1\rangle\cap \langle g_2\rangle=\{e\}$ because a cyclic group of order $5$ is generated by any of its nontrivial elements. We conclude that the number of elements of order $5$ is a multiple of four, hence is either $4$ or $8$ (as $0$ is excluded by the problem statement). This means that at least one element $g$ has order different from both $1$ and $5$. The order of $g$ must divide the order of the group by Lagrange, hence it is either $10$ and we are done, or it is $2$. Now if $h$ is of order $5$, then we note that $gh$ is of order $10$ because $(gh)^1\ne e$ (from $g=g^{-1}\ne h$) and $(gh)^2=h^2\ne e$ and $(gh)^5=g^5=g\ne e$.

share|cite|improve this answer
The fact that a group of even order contains an element of order $2$ can also be proven by a direct counting argument, which is in fact a common exercise in introductory courses. – Tobias Kildetoft Dec 31 '12 at 19:33

Let $g\in G$ be that element of order $5$, and $H=\langle g\rangle$ the subgroup generated by that element. Now $G/H$ is order $2$, and so there is an element $k\in G-H$ such that $(kH)^2=H$, or put another way, $k^2\in H$. If $k^2\neq 1$, then $k^2$ generates $H$, and thus $G$ is cyclic (generated by $k$). If $k^2=1$, then $G$ is cyclic, generated by $kg$.

share|cite|improve this answer
@MJD: $G$ is abelian (also, of index 2, ...). – user641 Dec 31 '12 at 20:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.