# Simple proof that group of order $2p$ with nontrivial center is cyclic?

We have a group $G$ of order $2p$, $p$ an odd prime, which has a nontrivial center. We wish to show the group is cyclic.

This is a simple exercise which can be instantly killed using heavy cannons. However, since I was asked about it by students which barely reached quotients, I wonder what simple ways are to prove this; I couldn't find any.

Thanks.

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Let $g$ be a nontrivial element of the center. If $g$ generates the whole group, we are done; otherwise $g$ generates a cyclic subgroup of order $2$ (resp. $p$). Hence $G/\langle g \rangle$ is a cyclic subgroup of order $p$ (resp. $2$). It follows that $G$ is generated by $g$ and one other element $h$ such that $h^p = g^k$ for some $k$ (resp. $h^2 = g^k$ for some $k$). Since $g$ is in the center, $gh = hg$. But this implies $h$ is also in the center. Hence $G$ is abelian.

If $h$ generates the whole group we are done. Otherwise $h$ has order $2$ (resp. $p$), since it has order dividing $2p$ and cannot have order $p$ (resp. $2$), and $G \simeq C_2 \times C_p$ as desired.

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This is a nice solution, but it involves what is more or less the proof that if G/Z(G) is cyclic then G is abelian, which I believe to be "too" nontrivial for students of that specific course. Of course, it just might be that the TA's gave an exercise which is too much... – Gadi A Dec 8 '10 at 18:34
@Gadi A: this is the only way I could see how to do it without Cauchy's theorem or something stronger. – Qiaochu Yuan Dec 8 '10 at 18:37
@Gadi If by saying "they barely reached quotients" you mean "they recently reached quotients" (rather than "they haven't reached quotients yet"), then to prove that $G/Z(G)$ cyclic $\Rightarrow$ $G$ abelian seems like a perfectly reasonable exercise to me. – Alex B. Dec 9 '10 at 12:43

Qiaochu's answer is simpler, but the following almost classifies all groups of order $2p$, so I'll leave it here as well.

Lemma 1: The order of an element divides order of the group. No new proof, but easy enough.

Lemma 2: There exists an element of order $p$. Can circumvent usual proof as follows: If there is an element of order $2p$ G is cyclic. If no element of order $p$ or $2p$ then $a^2=e$ for all $a$, so $G$ is abelian ($abab=e$, so $ab=b^{-1}a^{-1}=ba$). Then either use the fact that you know all abelian groups, or the fact that then $G$ is a vector space over the field $Z_2$, so has basis and so has order $2^k$.

Call this element $a$.

Lemma 3: Any subgroup of index 2 is normal. There are two cosets, and one of them is the subgroup itself. So right and left cosets coincide.

Applying this to the subgroup generated by $a$, for $b$ not a power of $a$, we have $bab^{-1}=a^m$. The group contains only $a^k$ and $ba^l$'s, which are all distinct. If $m=1$ they all commute and $ba$ generates the cyclic group. If $m\neq1$ there is no center. (The only step in classification left is to show that different $m \neq 1$ lead to the same group, which is not hard either.)

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This is a nice approach, but Lemma 2 is still somewhat nontrivial. – Qiaochu Yuan Dec 8 '10 at 18:40
Part of Tobias's answer simplifies the last part of Lemma 2 (2 commuting elements of order 2 generate a subgroup of order 4, contradiction). – Max Dec 9 '10 at 10:31

Actually, this can be done very simply using the following two facts, both of which follow from things the students should know at this point (though they may not have seen a centralizer before, it is not too hard to understand). First, any subgroup of prime index is maximal. This follows directly from Lagrange. Second, the center is never maximal. This follows by observing that if the center is not the entire group, let x be an element not in the center. Then the centralizer of x is not the entire group, since then x would be in the center. But clearly the center is contained in the centralizer of x and does not contain x, so the centralizer of x is a subgroup properly between the center and the entire group. Now it is easy to see that the group must be abelian, and then it is easy to show that it must in fact be cyclic (it has an element of order 2, which is simpler to show than Cauchy, and then it just needs to be shown that it cannot have 2 elemens of order 2, since then it would have a subgroup of order 4 contradicting Lagrange). (As a side note, Cauchy's theorem for abelian groups is fairly simple to show just from knowledge of the order of a product of subgroups, and in fact the general theorem then follows from the class equation)

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