5
$\begingroup$

This is Exercise 2.6.5 of F. M. Goodman's "Algebra: Abstract and Concrete". I want to check my proof.

Exercise 2.6.5: Show that a subgroup (of a group) is normal if and only if it is the union of conjugacy classes.

My Attempt:

Let $N$ be a subgroup. Then

$$\begin{align} N\text{ is normal }&\Leftrightarrow \forall g\in G, N=gNg^{-1} \\ &\Leftrightarrow \forall n\in N \forall g\in G\exists m_{n, g}\in N, n=gm_{n, g}g^{-1} \tag{1}\\ &\Leftrightarrow N=\bigcup_{n\in N}\underbrace{\bigcup_{g\in G}\left\{gm_{n, g}g^{-1}\right\}}_{\text{conjugacy class of }n}\tag{2} \\ &\Leftrightarrow N=\bigcup_{n\in N}[n], \end{align}$$ where $[n]$ is the conjugacy class of $n$.$\square$

Is this proof valid?

Thoughts:

I hope to make $(1)$ to $(2)$ (and back) more explicit.

Please help :)

$\endgroup$
  • 2
    $\begingroup$ It would be a better proof if it consisted of English sentences rather than just a wall of symbols. $\endgroup$ – Henning Makholm Mar 7 '17 at 11:20
  • $\begingroup$ @HenningMakholm True. $\endgroup$ – Shaun Mar 7 '17 at 11:21
  • $\begingroup$ @HenningMakholm I wanted to use "if and only if" statements all the way. Presenting it like this seemed most efficient. $\endgroup$ – Shaun Mar 7 '17 at 11:22
11
$\begingroup$

If $N$ is a normal subgroup of group $G$ and $n\in N$ then $gng^{-1}\in N$ for every $g\in G$ or equivalently $[n]\subseteq N$ where $[n]:=\{gng^{-1}\mid g\in G\}$ is the conjugacy class of $n$.

This tells us that: $$N=\bigcup_{n\in N}[n]$$ If conversely $N$ is a subgroup of group $G$ that satisfies $N=\bigcup_{n\in N}[n]$ then it is immediate that $gng^{-1}\in N$ for every $n\in N$ and $g\in G$, so the conclusion that $N$ is a normal subgroup is justified.

$\endgroup$
  • $\begingroup$ in your definition of normal subgroup you write $gng^{-1}$ gor every g$\in $N .i think it is not 'N' but 'G'.please check it out... $\endgroup$ – Cloud JR Sep 22 '18 at 7:46
  • $\begingroup$ Elegant proof +1 $\endgroup$ – Cloud JR Sep 22 '18 at 7:48
  • 1
    $\begingroup$ @CloudJR Thank you. You are right. I repaired now. $\endgroup$ – drhab Sep 22 '18 at 11:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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