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I'm not very good with English terms of group theory but here is the question :

$$\forall H\trianglelefteq G \rightarrow \exists H' \trianglelefteq G : {G\over H} \approx H'$$

is above statement always true? or should there be some other constraints?

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4  
Consider the additive group of integers. –  Karolis Juodelė Nov 3 '12 at 21:43
    
@KarolisJuodelė what about it? I mean for whatever $\Bbb{Z}_n$ you choose there is always $n\Bbb{Z}$ to satisfy above guess and vise versa, and I honestly can't think of any other subgroup of $\Bbb{Z}$. –  Ali.S Nov 3 '12 at 21:47
3  
$n\mathbb{Z}$ is not isomorphic to $\mathbb{Z}_n$. –  wj32 Nov 3 '12 at 21:49
    
@wj32 it's not supposed to, but $n\Bbb{Z}$ is isomorphic to ${\Bbb{Z}\over\Bbb{Z}_n}$ –  Ali.S Nov 3 '12 at 21:50
3  
@Gajoo: $\mathbb{Z}_n$ is not even a subgroup of $\mathbb{Z}$! –  wj32 Nov 3 '12 at 21:52

2 Answers 2

up vote 3 down vote accepted

This is not true in general. The smallest counterexample can be found in the quaternion group $Q_8$. There $Q_8/Z(Q_8)$ is isomorphic to the Klein $4$-group $\mathbb{Z}_2 \times \mathbb{Z}_2$, but every subgroup of order $4$ in $Q_8$ is cyclic.

However, if we assume that $G$ is finite and abelian, then the statement is true.

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1  
Um, the only nontrivial quotient of $S_3$ is of order two, but the order-two subgroups there are nonnormal. –  Lubin Nov 3 '12 at 22:35
    
@Lubin: Ah, you're right, I forgot we wanted a normal subgroup here.. –  Mikko Korhonen Nov 3 '12 at 22:36
    
I think this is still a strong example, because it shows that $G/H$ not only does not have to be isomorphic to a normal subgroup of $G$, but $G/H$ does not have to be isomorphic to a subgroup of $G$ at all (a common mistake). –  Alexander Gruber Nov 3 '12 at 22:50

Hint: The symmetric group $S_5$ has exactly three normal subgroups: $\{1\}$, $S_5$ and the alternating group $A_5$, which has index 2 in $S_5$.

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All you really need here is that $A_5$ has index $2$ in $S_5$, since normal subgroups of order $2$ are always central and $S_5$ has trivial center. –  Mikko Korhonen Nov 3 '12 at 22:06
    
What on earth is there to downvote about this? –  rschwieb Nov 5 '12 at 11:53

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