I'm having difficulties finding a proof for the following problem:

"Let $p$ be a prime and $G$ a group of order $2p$ which contains a normal subgroup $N$ of order $2$. Show that $G$ is an Abelian Group."

My approach would have been the following: The factor group $G/N$ has order $p$. As $p$ is prime, $G/N$ is a cyclic group, in particular Abelian. Thus $xyN = yxN$ $\forall x,y \in G$.
Furthermore $N = \{e,a\}$ ($e$ is neutral element) for an $a \in G$. Thus $a^{-1} = a$. Thus we can assert that $ag = ga$ $\forall g \in G$. Now let $x,y \in G$, show that $xy = yx$. We know that $xye \in yxN$, so either $xye = yxe$ (and we're done) or $xye = yxa$. Now this is where I'm stuck. I wanted to show that $xye = yxa$ cannot be, but this is not as obvious as I had thought.

Do you have any hints? Is my approach misleading?

Thanks a lot in advance for any help!

  • $\begingroup$ why is $ag=ga$ for all $g$? $\endgroup$
    – janmarqz
    Oct 24, 2015 at 14:25
  • 2
    $\begingroup$ $gag^{-1}$ is element of $N$. If $gag^{-1} = e$, $g^{-1} = ag^{-1}$, so $a = e$ in contradiction to $N$ order of 2. So $gag^{-1} = a$ and $ga = ag$. $\endgroup$
    – johnnycrab
    Oct 24, 2015 at 14:49
  • 1
    $\begingroup$ ok, nice. We can simplify as: if $gag^{-1}=e$ then $ga=g$, thus $a=e$. $\endgroup$
    – janmarqz
    Oct 24, 2015 at 14:55

3 Answers 3


We can assume that $p$ is an odd prime. If $H$ is a subgroup of order $p$ (which exists, by Cauchy's theorem), it has index $2$ in $G$, so it must be normal. Indeed, if we write $G=H\sqcup gH$ then we cannot have $H=Hg$ for $1g=g\notin H$ by the first coset decomposition, so $G=H\sqcup Hg$ and $Hg=gH$. This works for any $g\notin H$, and is of course valid if $g\in H$, so $H$ is normal.

If $K$ is a normal subgroup of order two it has in particular order coprime to that of $H$, hence $H\cap K=1$. By cardinality considerations it follows that $HK=G$, and if $K$ is normal it follows that $G$ is the direct product of $H$ and $K$, so $G$ is isomorphic to $C_2\times C_p$, which is manifestly abelian. The argument above shows that any group of order $2p$ is isomorphic to either $C_2\times C_p=C_{2p}$ or to $C_2\ltimes C_p=D_{2p}$ the dihedral group of order $2p$. Note that any nontrivial semidirect product as above must be $D_{2p}$. Indeed, any nontrivial automorphism of order $2$ of $C_p$ sends a generator $r$ to some $r^i$ with $r^{i^2}=r$ (apply the automorphism again), or what is the same, $i^2=1\mod p$. This means that $i=1\mod p$ (which cannot happen since the identity has order $1$) or $i=-1\mod p$; that is $r\mapsto r^{-1}$, so the only nontrivial automorphism of $C_p$ of ordern $2$ is inversion.

The argument above can be modified as follows. Take $s$ an element of order $2$, $r$ an element of order $p$ in $G$, by Cauchy. Since the subgroup generated by $r$ is normal, $srs^{-1}=srs=r^i$ for some $i$, and raising to the $i$-th power we see that $sr^is= r^{i^2}$, but since $s^2=1$ we get $r^{i^2}=r$. Since $r$ has order $p$; it follows that $i^2=1\mod p$. If $i=1 \mod p$, $G$ is abelian for $srs^{-1}=r$, i.e. $sr=rs$, that is $r,s$ commute (and they generate all of $G$). Else we see that $i=-1\mod p$, so we get indeed $srs=r^{-1}$, which is the relation we wanted.

  • $\begingroup$ $H$ is already a normal subgroup so I don't see where you used the fact that $K$ is normal without using $H$? I was wondering if the subgroup of order $2$ can be normal ? $\endgroup$
    – JeSuis
    Oct 26, 2015 at 12:02
  • $\begingroup$ @user281591 I characterized groups of order $2p$, you can deduce your result from the above. Read it carefully. $\endgroup$
    – Pedro
    Oct 26, 2015 at 13:48
  • $\begingroup$ You didn't read my comment carefully, I know this characterization but in the proof we don't need the assumption that the group of order $2$ is normal only the fact that the group of order $p$ is normal ( which is a consequence). So my question was can we prove that the group of order $2$ is normal? $\endgroup$
    – JeSuis
    Oct 26, 2015 at 14:53
  • $\begingroup$ @user281591 We cannot, because it sometimes isn't normal. When it is, your group is $C_{2p}$, when it isn't, your group is $D_{2p}$. $\endgroup$
    – Pedro
    Oct 26, 2015 at 14:56
  • $\begingroup$ I don't see why if the group is normal then $G$ is $C_2p$, for me it's $C_2p$ when we assume that the $G$ is cyclic, otherwise it's $D_2p.$ $\endgroup$
    – JeSuis
    Oct 26, 2015 at 15:01

Suppose $G \neq Z(G).$ If $N$ is normal, then $N=Z(G)$. But $G/Z(G) $ cyclic $\Rightarrow$ G abelian.


I think that you can prove that a group G is abelian only proving that all its subgroups are normal. By Cauchy's theorem apply to the prime p (and later to the prime 2) you have only one p-subgroup and only one a 2-subgroup. If only exist one p-sylow you have a result that say it's normal. So you have the result following the Lagrange formula (you have only subgroups of orders 2 or p and in any case they are normal)

  • 3
    $\begingroup$ Are you saying that $\text{all subgroups of $G$ are normal} \implies G \text{ is abelian}?$ This is not true, the quaternion group $Q_8$ has only normal subgroups. $\endgroup$
    – pjs36
    Oct 24, 2015 at 22:04
  • $\begingroup$ Yes you are all rigth, it was false my assertion, I'm very sorry. I think that in this particular case I can fix it: if you name N,M (subgroups of order 2 and p) that are normal by the last argue, i think you can prove that NM=MN only argue by the orders because 2 and p are prime numbers but i'm not sure. In any case i think that argue by sylow and cauchy theorems it's the easy attach $\endgroup$
    – Patricio
    Oct 24, 2015 at 22:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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