Take the 2-minute tour ×

I am thinking on the following problem:

If $|G|=12$ and there is no element of order $2$ in its center then $3$-Sylow subgroup of $G$ cannot be normal in $G$.

I was told that to assume the $3$-Sylow subgroup of $G$, say $P$, is normal in the group and go to reach a contradiction.

My attept: $|P|=3$ so it is cyclic, $P=\langle x\rangle=\{1,x,x^2\}$. As above hint, I would have for all $g\in G$ two possibilities: $gxg^{-1}=x$ or $gxg^{-1}=x^2$. Cannot to go any further. I am in doubt if the hint leads me to a contradiction properly and if so, it does with a long proof. Any help or any other way are welcome to me. Thanks for your time.

share|improve this question
This is a followup to math.stackexchange.com/questions/158339/… –  Jack Schmidt Jun 24 '12 at 19:10

2 Answers 2

up vote 4 down vote accepted

Hint #1: Show that all Sylow 2-subgroups are abelian. So if an element of a Sylow 2-subgroup commutes with the elements of $P$, then it belongs to the center.

Hint #2: At this point the questions you should ask from yourself are: 1) What possibilities (up to isomorphism) are there for the Sylow 2-subgroup H? 2) Given that $P$ was assumed to be normal, what possibilities are there for the conjugation action for the non-trivial elements of $H$. Remember that conjugation action is a homomorphism from $H$ to $Aut(P)$.

share|improve this answer
:About you first hint: That the Sylow 2-subgroups is abelian is obvious cause they are of order 4. So I should show that an element of one of these subgroups COMMUTE with $x$. Thanks. As you said me later, you put a loop in my mind finding the result. :-) –  Babak S. Jun 24 '12 at 10:14
Correct (re: the 1st hint). Be a bit careful in that the element of $H$ that commutes with $P$ should be of order two. It doesn't really matter much in this case, but for full credit... :-) –  Jyrki Lahtonen Jun 24 '12 at 10:29
I am on the way you paved. Yes, with that element say $y$, I can see that $G$ is being generated by $x$ and $y$. Thanks. Thanks. –  Babak S. Jun 24 '12 at 10:38

To follow your attempt:

For every $g \in G$, we have $gxg^{-1} = x$ or $gxg^{-1} = x^2$. Thus $x$ has $1$ or $2$ conjugates, in other words, $[G : C_G(x)] = 1$ or $[G : C_G(x)] = 2$. In both cases, $2$ divides the order of $C_G(x)$, and thus $C_G(x)$ contains an element $y$ of order $2$. This gives us the desired contradiction. Can you see why $y$ must belong to the center of $G$?

share|improve this answer
Thanks for your answer. It solves my problem in detail. :) –  Babak S. Jun 25 '12 at 5:39

Your Answer


