My Math Forum H < Z(G) and G/H nilpotent, implies G is nilpotent.

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 June 23rd, 2016, 05:13 AM #1 Member   Joined: Oct 2013 Posts: 36 Thanks: 1 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.
 June 25th, 2016, 03:58 AM #2 Senior Member   Joined: Sep 2008 Posts: 150 Thanks: 5 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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