MHB Is $\theta(N)$ a normal subgroup of $G$?

[Ackbach]

Here is this week's POTW:

-----

Let $G$ be a group. If $\theta$ is an automorphism of $G$ and $N \vartriangleleft G$, prove that $\theta(N) \vartriangleleft G$.

-----

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Ackbach]

This was Problem 2.5.27 on page 75 of I. N. Herstein's Abstract Algebra, 3rd Ed..

No one answered this week's POTW. The solution follows:

[sp]
By definition, $\theta$ is 1-1, onto, a homomorphism, and $\theta: G \to G$. By definition, since $N \vartriangleleft G$, it must be that $N$ is a subgroup of $G$, and $a^{-1}Na\in G$ for every $a\in G$. We first show that $\theta(N)$ is a subgroup of $G$. Since $\theta:G\to G$, and $N\subseteq G$, it follows that $\theta(N)\subseteq G$. Since $N$ is a subgroup, it follows that $e\in N$, and hence $\theta(e)=e\in G$, since homomorphisms map identities to identities (Lemma 2.5.2 in Herstein). Therefore, $\theta(N)\not=\varnothing$. Suppose $s, t\in\theta(N)$. By definition, there exist $m, n\in N$ such that $s=\theta(m)$ and $t=\theta(n)$. Then $st=\theta(m)\theta(n)=\theta(mn)$. Since $N$ is a subgroup, $mn\in N$, hence $st\in\theta(N)$, and $\theta(N)$ is closed under multiplication. Moreover, since $N$ is a subgroup, $m^{-1}\in N$, and hence $\theta(m^{-1})=\theta(m)^{-1}=s^{-1}\in\theta(N)$, since homomorphisms preserve inverses (Lemma 2.5.2 in Herstein, again). Since $\theta(N)$ is a nonempty subset of $G$, and is closed under multiplication and inverses, it is a subgroup.

Now let $a\in G$ be arbitrary, as well as $u\in\theta(N)$ be arbitrary. Then we wish to show that $aua^{-1}\in \theta(N)$. There exists $p\in N$ such that $u=\theta(p)$. Therefore, we have that $aua^{-1}=a \, \theta(p) \, a^{-1}=\theta(apa^{-1}).$ Since $N \vartriangleleft G$, it follows that $apa^{-1}\in N$, hence $\theta(apa^{-1}) \in \theta(N)$, and $\theta(N)\vartriangleleft G$, as required.
[/sp]