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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.

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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: $|xy|=10$

Proof:

$(xy)^{|xy|}=e$, 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)

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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
4  
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 $(gh)^5=g^5=g\ne e$.

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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 $kg$.

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

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