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Problem of the Week # 252 - Feb 07, 2017

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Ackbach

Indicium Physicus
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Jan 26, 2012
4,202
Here is this week's POTW:

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

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,202
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:


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.
 
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