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June 23rd, 2016, 05:13 AM   #1
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H < Z(G) and G/H nilpotent, implies G is nilpotent.

So I am doing this question from Rotman (pg 117, q5.38 ): if H is a subgroup of the center of G and G/H is nilpotent then G is nilpotent.

I have looked at other proofs on other websites but they do not match what I have down so I just wanted to check if what I did was right too.

I said let $|G| = p_1^{e_1}...p_n^{e_n}|H|, p_i$ are primes.
Then $G/H \cong P_1 \times ... \times P_n$ where $P_i$ is a Sylow $p_i$-subgroup of order $p_i^{e_i}$ since $G/H$ is nilpotent.
Now since $H$ is abelian it is nilpotent too but also we can get an isomorphism from $Q_1 \times ... \times Q_n \times H$ to $G$ where $Q_i$ is the preimage of $P_i$ in $G$ by the natural map.
Then since $Q_1 \times ... \times Q_n$ and $H$ are both nilpotent so is $G$ as their direct product.

Does this work?

Edit: damn realised this only works for finite G- in any case is this still valid if G is finite?

Last edited by fromage; June 23rd, 2016 at 05:22 AM.
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June 25th, 2016, 03:58 AM   #2
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The line, were you state the isomorphism of G with a Group that has H as a factor is wrong: A counterexample would be G cyclic of order 4 and H the unique subgroup of order 2.

I would use the characterization of nilpotent with the Central series to prove your claim.
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