Does there exist a finite group G with a normal subgroup H, for which G has no subgroup isomorphic to the quotient group G/H?

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Andrei Jaikin

Yes. Consider for example $G=SL_2(F_p)$ for $p\ge 5$ and $H=Z(G)$.

Błażej Szepietowski

Yes, for example, the quaternion group Q has a unique subgroup H={1,-1} of order 2, for which Q/H is the Klein four-group.

Jie Wang

The answer is YES. Let G=SL(2,3). Then |G|=24 and G has a normal subgroup H of order 2. So the quotient group G/H is a group of order 12. However, G has no subgroup of that order.

Saeid Jafari

Thanks a lot for your answers.

So the structure of a finite group really depends on the chief factors, not only the subgroups.

Now a dual question arise, that is " Does there exist a group G with a nontrivial proper subgroup K, such that K is not isomorphic to a quotient of G? "

but about this question obviously we can put G=A_5 , the simple alternating group

and let K to be a subgroup of order 2. Then G has no a quotient isomorphic to K.

Eka Ratna Acharya

Perhaps we may check it from Lagrange's Theorem for groups!

Ram parkash Sharma

In symmetric group S_3 , there is a normal subgroup K={e, (123), (132)} of order 3. But there is no normal subgroup H of order 2, so K is not isomorphic to a quotient of G.