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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 ?

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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 $Z(A \times B) = Z(A)\times Z(B)$

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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 $(i, 0)$.

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