# Abstract Nonsense

## Review of Group Theory: Semidirect Products (Pt. II)

Point of post: This is a continuation of this post.

It’s interesting then to see exactly what the semidirect product “means” in terms of subgroups. By this, I mean that the direct product of groups can be thought of as just the internal product of subgroups, but there was some cosmic accident and they actually aren’t subsets of any coherent group. The question then remains how we can interpret this for semidirect products. We begin by noticing something interesting about certain subgroups of $G\rtimes_\varphi H$ and use this to characterize the semidirect product

Theorem: Let $G$ and $H$ be groups and $\varphi:H\longrightarrow \text{Aut}\left(G\right)$. Then, if $\tilde{G}$ and $\tilde{H}$ are defined as before then for all $(g,h)\in G\times H$:

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\begin{aligned}&\mathbf{(1)}\quad \tilde{G}\unlhd G\rtimes_\varphi H\text{ and }\left(G\rtimes_\varphi H\right)/\tilde{G}\cong H\\ &\mathbf{(2)}\quad H\leqslant G\rtimes_\varphi H\\ &\mathbf{(3)}\quad \tilde{G}\cap\tilde{H}=\{e\}\\ &\mathbf{(4)}\quad G=\tilde{G}\tilde{H}\\ &\mathbf{(5)}\quad (e,h)(g,e)(e,h)^{-1}=\left(\varphi_h(g),e\right)\end{aligned}

Proof:

$\mathbf{(1)}$: Define

$\pi:G\rtimes_\varphi H\to H:(g,h)\mapsto h$

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This is evidently surjective and a homomorphism. Indeed:

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$\displaystyle \pi\left((g,h)(g',h')\right)=\pi\left(\left(g\varphi_h(g'),hh'\right)\right)=hh'=\pi\left((g,h)\right)\pi\left((g',h')\right)$

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Noting though that

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$\displaystyle \ker\pi=\left\{(g,h)\in G\times H:h=e\right\}=\tilde{G}$

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enables to conclude, from the First Isomorphism Theorem, that $\tilde{G}\unlhd G\rtimes_\varphi H$ and $\left(G\rtimes_\varphi H\right)/\tilde{G}\cong H$.

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$\mathbf{(2)}$: This follows from the fact that if $(e,h),(e,h')\in \tilde{H}$ then

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\begin{aligned}(e,h)(e,h')^{-1} &= (e,h)(\varphi_{h^{-1}}(e),h^{-1})\\ &= (e,h)(e,h'^{-1})\\ &= (e\varphi_h(e),hh'^{-1})\\ &= (e,hh'^{-1})\in \tilde{H}\end{aligned}

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$\mathbf{(3)}$: This is immediate.

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$\mathbf{(4)}$: Let $(g,h)\in G\rtimes_\varphi H$ be arbitrary. Then, We merely note that

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$(g,e)(e,h)=\left(g\varphi_e(e),eh\right)=(g,h)$

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from where the result is clear.

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$\mathbf{(5)}$: This is just a simple computation:

\begin{aligned}(e,h)(g,e)(e,h)^{-1} &= (e,h)(g,e)(\varphi_{h^{-1}}(e),h^{-1})\\ &= (e,h)(g,e)(e,h^{-1})\\ &= (e,h)(g\varphi_e(e),eh^{-1})\\ &= (e,h)(g,h^{-1})\\ &= (e\varphi_{h}(g),hh^{-1})\\ &= (\varphi_h(g),e)\end{aligned}

$\blacksquare$

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In fact, the first four of the above conditions characterize the semidirect product in the following sense:

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Theorem: Let $G$ be a group with $N\unlhd G$ and $K\leqslant G$. Moreover, if $N\cap K=\{e\}$ and $G=NK$ then $G\cong N\rtimes_\varphi K$ where

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$\varphi:K\longrightarrow \text{Aut}\left(N\right):h\mapsto i_h$

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where $i_h$ is a variant of the inner automorphism $i_h(n)=hnh^{-1}$. (note that this makes sense that $i_h\in\text{Aut}(N)$ since $N$ is normal and so $i_h\left(N\right)=N$).

Proof: It’s evident from the conditions that $G=NK$ and $N\cap K=\{e\}$ that every element $g$ of $G$ may be writtenuniquely as the product $nk$ with $n\in N$ and $k\in K$. Indeed, by the fact that $G=NK$ we have that every element $g$ of $G$ may be written as the product of $n\in N$ and $k\in K$. Note though that if $nk=g=n'k'$ then $k'^{-1}k=n^{-1}n'$ and thus $n^{-1}n'\in N\cap K$ so that $n^{-1}n'=e$ or $n=n'$ from where it quickly follows that $k=k'$ and thus uniqueness is established. Thus, the mapping

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$\displaystyle \Phi:N\rtimes_\varphi K\to G:(n,k)\mapsto nk$

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is a bijection. Note though that

\displaystyle \begin{aligned}\Phi\left((n,k)(n',k')\right) &= \Phi\left(\left(n\varphi_k(n'),kk'\right)\right)\\ &= \Phi\left(\left(nkn'k^{-1},kk'\right)\right)\\ &= nkn'k^{-1}kk'\\ &= nkn'k'\\ &= \Phi\left((n,k)\right)\Phi\left((n',k')\right)\end{aligned}

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from where it follows that $\Phi$ is a homomorphism and thus $N\rtimes_\varphi K\cong G$ as desired. $\blacksquare$

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References:

1. Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Mathematical Society, 1996. Print.

2. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 200

3.Grillet, Pierre A. Abstract Algebra. New York: Springer, 2007. Print.