Suppose, that $ K $ is normal subgroup of $ N $, and $ N $ is also normal subgroup of $ G $. Prove, that $ K $ needn't be normal subgroup of $ G $. I can give only counter-example?
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2"mustn't" makes the statement incorrect. You mean "needn't". – Zev Chonoles Aug 20 '13 at 21:29
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Although the question isn't very well posed, you are being asked to produce groups $K \lhd N \lhd G$ with $K \not \lhd G.$ – Geoff Robinson Aug 20 '13 at 21:31
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@ZevChonoles thank you. Ok, so if I give counter-example it will be enough? – Mat Aug 20 '13 at 21:33
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@Mat The proposed duplicate question gives more details than I would have given. You are lucky. – Mark Bennet Aug 20 '13 at 21:37
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@MarkBennet But I had counter-example :), I asked only about if this enough :) – Mat Aug 20 '13 at 21:38
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And unfortunately I haven't found this topic before, so I must only asked :). – Mat Aug 20 '13 at 21:40
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If you can produce an example where $K$ is not normal in $G$ then you have proved what you were asked to prove. – Mark Bennet Aug 21 '13 at 03:55
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Look at the group $A_4$ - the alternating group on four elements. This has a normal subgroup $V$ of order $4$ which is abelian. $V$ has three subgroups of order $2$, all of which are normal (since $V$ is abelian). But $A_4$ has no normal subgroup of order 2.
I'll leave you to fill in the details, since you have given no indication of your own effort on the problem.
Mark Bennet
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