Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm trying to prove that the center of $A_n\times \mathbb{Z}/2 \Bbb Z $ is nontrivial. I read that it contains an element of order 2 that commutes with everything, but I don't know how to proof this. Can anybody help me out here ?

share|improve this question

2 Answers 2

up vote 2 down vote accepted

I'm far from being sure but:

Take $f$ the neutral element of $A_n$

$\forall a \in A_n, af = a = fa$

$\forall b\in \Bbb Z / 2 \Bbb Z, b+1=1+b$ because $\Bbb Z / 2 \Bbb Z$ is abelian

So $\forall (a,b)\in A_n \times \Bbb Z / 2 \Bbb Z, (f,1)(a,b)=(a,b)(f,1)$, that is $(f,1)\in Z(A_n \times \Bbb Z / 2 \Bbb Z)$

$(f,1)\not= e = (f,0)$ so $Z(A_n \times \Bbb Z / 2 \Bbb Z)$ isn't trivial

I haven't studied groups but there must be some kind of property like $Z(A \times B) = Z(A)\times Z(B)$

share|improve this answer
Very nice proof for someone who "hasn't studied groups"....How then did you know how is the direct product of groups defined?! –  DonAntonio Apr 16 '13 at 22:50
The final property noted here is un-necessary in answering this question. –  Benjamin Dickman Apr 16 '13 at 22:52
Well I've studied the definition and a few basic properties. It really was presented just as a stepping stone to vector spaces and to be able to put a structure on $GL_n(\Bbb R)$ and its subgroups. But the only things I know (or guess) about the center etc. are things I've seen here on Stackexchange :) –  xavierm02 Apr 16 '13 at 22:53
@B.D : No but it shows that as long as one of the groups has a non-trivial center, the center of the product is non-trivial. –  xavierm02 Apr 16 '13 at 22:56
Thanks xaviermo2! –  Kasper Apr 16 '13 at 23:01

The element $(i, 1)$ where $i \in A_n$ is the identity is an element of the center of the crossed groups.

It is also different from $e$, by which I am sure you mean $(i, 0)$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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