N'$ implies $N\cong N'$ for the normal subgroups?

Asked Jun 25 '15 at 11:24

Active Jun 25 '15 at 11:24

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We have normal subgroups $ N$ and $ N'$ of a group $ G $. Is the following statement correct?

$G/N\cong G/N'\Leftrightarrow N\cong N'$.

$\Leftarrow $ is trivial.

$\Rightarrow:$ I don't know how to start. Or should I search for counter examples?

I would be pleased, if someone could give a hint.

asked Jun 25 '15 at 11:24

Daniel Bernoulli

2

Neither direction is true, so you should probably think a bit more about the "trivial". – Tobias Kildetoft Jun 25 '15 at 11:25
Also, I am certain this is a duplicate, but I can't find to what right now. – Tobias Kildetoft Jun 25 '15 at 11:26
Related: Finite group with isomorphic normal subgroups and non-isomorphic quotients? and Isomorphic quotients by isomorphic normal subgroups and several linked questions. – Bart Michels Jun 25 '15 at 11:27
3

An example to what @TobiasKildetoft said. Take $$G=C_4\times C_2=\langle x\rangle\times \langle y \rangle.$$ Now take $N_1=\langle y \rangle$ and $N_2=\langle x^2\rangle$. – Ofir Schnabel Jun 25 '15 at 11:30
Thank you for the links @barto. I could not find them, when I wrote the article. For the "trivial" case, I only thought of equality and not isomorphy. Great mistake! Sorry! – Daniel Bernoulli Jun 25 '15 at 12:01

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