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Here is my problem:

Let $$H=\langle a,b| a=bab, b=aba\rangle $$ and $\frac{H}{A}\cong\ Q_8$ wherein $A\leq Z(H)\cap H'$. Show that $H\cong Q_8$.


Working on the elements, I could see that $H\cong Q_8=\langle a,b| a^4=1, a^2=b^2,ab=ab^3\rangle$. In fact, the $G$'s relations can be yielded by $H$'s.

Yet; I don't see the additional points, above problem wanted to give me? Thanks.

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The only possibility for $A$ seems to be $A\cong\{+1\}$ and $Z(H)\cap H^\prime\cong\{-1,+1\}$. –  Elias Dec 15 '12 at 15:53
    
@Elias: It sems to me that if I could show $H'\cong\mathbb Z_2$, everything would be correct. This is what you noted first. But this is itself a new problem. :) –  B. S. Dec 15 '12 at 16:13
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There is in fact a general result that says that if $H$ if $G$ is any finite group with a presentation having the same number of generators and relations, then the Schur Multiplier of $G$ is trivial. This is equivalent to the statement that, if $H/A \cong G$ with $A \le Z(H) \cap H'$ then $A=1$. –  Derek Holt Dec 15 '12 at 18:38
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@DerekHolt: OMG! I saw it when I did some handy calculations on $Q_n$. I didn't notice this point. Thanks.Thanks for lighting me. –  B. S. Dec 15 '12 at 18:48
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If the finite group $G$ has a presentation with $n$ generators and $n$ relations, then $G=F/R$ with $F$ free of rank $n$ and $R$ the normal closure of $n$ elements of $F$. So $R/[R,F]$ is an $n$-generated abelian group. But its quotient group $R/(R \cap F')$ has finite index in the free rank $n$ abelian group $F/F'$, so $R/(R \cap F')$ is free abelian of rank $n$ and hence so is $R/[R,F]$. So its torsion subgroup $(R \cap F')/[R,F]$, which is the Schur Multiplier of $G$, is trivial. –  Derek Holt Dec 16 '12 at 10:49
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