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Can you show that $\,exp(G/Z(G))=2 \Longrightarrow exp(G')=2$ ?

My try is: clearly $\,G/Z(G)\,$ is abelian group,so G is nilpotent of class 2 and we have [G',G]=1 if x ϵ G' then [x,g]=1 ∀ g ϵ G so G'≤Z(G).

But I can't show that O(x)=2

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Ok, this time I think I understood: first, it isn't true that $\,G/Z(G)\,$ has to be abelian (for an easy counter example take $\,G=S_3\,$) , so what are your assumptions and what is your question? – DonAntonio Oct 28 '12 at 17:53
my assumptions is exp(G/Z(G))=2 Donatonio not that exp(S_3)=6 – mojtaba farazi Oct 28 '12 at 18:18
Ok. So you assume that $\,G/Z(G)\,$ is abelian of exponent 2? And then, of course, $\,G\,$ is nilpotent of class 2. What do you mean by $\,O(x)\,$? – DonAntonio Oct 28 '12 at 18:57

Lemma Let $G$ be a group and $x_1, x_2, ... x_n \in G$, with $x_1\cdot x_2 \cdot ... \cdot x_n \in Z(G)$. Then $x_1\cdot x_2 \cdot ... \cdot x_n =x_2\cdot x_3 \cdot ... \cdot x_n \cdot x_1$ (in other words, the elements in the product can be permuted cyclically).
Proof. Since $x_1\cdot x_2 \cdot ... \cdot x_n \in Z(G)$, $x_1^{-1} \cdot x_1\cdot x_2 \cdot ... \cdot x_n \cdot x_1 = x_1\cdot x_2 \cdot ... \cdot x_n \square$.

It was already observed that $G' \subset Z(G)$, so by the lemma $[x,y]= x^{-1}y^{-1}xy= y^{-1}xyx^{-1}$. Because $exp(G/Z(G))=2$, it follows that $x^2 \in Z(G)$, say $x^2 = z$, whence $x = x^{-1}z$, with $z \in Z(G)$. Now substitute : $y^{-1}xyx^{-1} = y^{-1}(x^{-1}z)y(xz^{-1}) = y^{-1}x^{-1}yx$, since $z \in Z(G)$. We now have that $[x,y]= [y,x] = [x,y]^{-1}$. We conclude that $exp(G')=2$.

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Thanks for the answer. – mojtaba farazi Oct 29 '12 at 11:56
Let G be a group such that Z(G/Z(G))=3.How do I demonstrate G is nilpotent with CL(G)≤4? – mojtaba farazi Oct 29 '12 at 17:31
Mojtaba, probably you are new to this site, but please mark my answer as the answer you were looking for. Thanks. – Nicky Hekster Oct 29 '12 at 19:19
OK sorry I am new to site – mojtaba farazi Oct 29 '12 at 19:29
@mojtabafarazi: Neat answer, Nicky +1 – Babak S. Oct 31 '12 at 6:18

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