Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question

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.

share|cite|improve this answer

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.

share|cite|improve this answer
But how do we prove $H$ is a characteristic subgroup of $N$ ? – rTeja Nov 22 '12 at 4:51

Yes. This is one of the several cases when normality is transitive. If

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

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

share|cite|improve this answer
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 '14 at 0:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.