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!

  • why is $ag=ga$ for all $g$? – janmarqz Oct 24 '15 at 14:25
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    $gag^{-1}$ 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 $ga = ag$. – johnnycrab Oct 24 '15 at 14:49
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    ok, nice. We can simplify as: if $gag^{-1}=e$ then $ga=g$, thus $a=e$. – janmarqz Oct 24 '15 at 14:55
up vote 2 down vote accepted

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.

  • $H$ 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 ? – JeSuis Oct 26 '15 at 12:02
  • @user281591 I characterized groups of order $2p$, you can deduce your result from the above. Read it carefully. – Pedro Tamaroff Oct 26 '15 at 13:48
  • 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? – JeSuis Oct 26 '15 at 14:53
  • @user281591 We cannot, because it sometimes isn't normal. When it is, your group is $C_{2p}$, when it isn't, your group is $D_{2p}$. – Pedro Tamaroff Oct 26 '15 at 14:56
  • 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 $D_2p.$ – JeSuis Oct 26 '15 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)

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    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. – pjs36 Oct 24 '15 at 22:04
  • 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 – Patricio Oct 24 '15 at 22:11

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