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Is it true that Subgroup of a Cyclic Normal subgroup of a Group is again Normal ? If so any hints for the proof?

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3 Answers

Sure. It follows from a more general fact: a characteristic subgroup of a normal subgroup of $G$ is also a normal subgroup of $G$. It's even easier to think about the question in these general terms.

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If $H<N\lhd G$ and $N$ is cyclic, then $H$ is a characteristic subgroup of $N$, i.e. it is left invariant by all automorphism of $N$, thus including those outer automorphisms of $N$ that are actually inner automorhisms of $G$, i.e. $H$ is conjugation invariant.

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But how do we prove $H$ is a characteristic subgroup of $N$ ? –  rTeja Nov 22 '12 at 4:51
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Yes. This is one of the several cases when normality is transitive. If

$$N\triangleleft C\triangleleft G\,\,\,,\,\,C=\text{cyclic, then}\;\; C=\langle c\rangle\Longrightarrow N=\langle c^k\rangle\Longrightarrow$$

$$\Longrightarrow \,\,\forall g\in G\,\,,\,\,x^{-1}c^{rk}x=(x^{-1}c^kx)^r\in N\,....$$

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How can we say that $x^{-1}c^kx \in N$ ? –  rTeja Nov 22 '12 at 4:52
    
I believe it would have been clearer to use $(x^{-1}c^rx)^k \in N $ in the last line. –  Quester Mar 1 at 0:49
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