# Abstract Nonsense

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

Point of post: In this post we discuss the fruitful notion of semidirect products.

Motivation

Often the direct product of groups isn’t general enough to deal with most purposes. Thus, the idea of the semidirect product is to capture the spirit of what the direct product does but do so in a more general manner which with the suitable choice of homomorphis (see below) reduces to the case of the direct product anyways.

Semidirect Product

Let $G$ and $H$ be groups and $\varphi$ a homomorphism $H\to\text{Aut}\left(G\right):h\mapsto \varphi_h$ where $\text{Aut}(H)$ is, as always, the group of automorphisms on $H$. Then we define a group structure on the set $G\times H$ by the rule

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

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we call this group structure on $G\times H$ the semidirect product of $G$ and $H$ with respect to $\varphi$ and denote it $G\rtimes_\varphi H$. We first prove (since this is much less apparent then the case of the direct product) that this forms a group structure:

Theorem: If $G$ and $H$ are groups and $\varphi:H\longrightarrow \text{Aut}(G)$ is a homomorphism then $G\rtimes_\varphi H$ under the operations defined above is a group with identity $(e_G,e_H)$ and inverses given by

$(g,h)^{-1}=\left(\varphi_{h^{-1}}(g^{-1}), h^{-1}\right)$

Proof: To see that this operation is associative we note that

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\begin{aligned}(g_1,h_1)\left((g_2,h_2)(g_3,h_3)\right) &= (g_1,h_1)\left(g_2\varphi_{h_2}(g_3),h_2h_3\right)\\ &= \left(g_1\varphi_{h_1}\left(g_2\varphi_{h_2}(g_3)\right),h_1(h_2h_3)\right)\\ &= \left(g_1\varphi_{h_1}(g_2)\varphi_{h_1}\left(\varphi_{h_2}(g_3)\right),(h_1h_2)h_3\right)\\ &= \left(g_1\varphi_{h_1}(g_2)\varphi_{h_1h_2}(g_3),(h_1h_2)h_3\right)\\ &= \left(g_1,\varphi_{h_1}(g_2),h_1,h_2\right)\left(g_3,h_3\right)\\ &= \left((g_1,h_1)(g_2,h_2)\right)(g_3,h_3)\end{aligned}

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where we’ve made use that $\varphi:H\to\text{Aut}(G)$ is a homomorphism. To prove that $(e_G,e_H)$ acts as a two-sided identity we note that

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$\displaystyle (e_G,e_H)(g,h)=(e_G\varphi_{e_H}(g),e_Hh)=(g,h)$

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(where we’ve made use of the fact that since $\varphi:H\to\text{Aut}(G)$ is a homomorphism we have that $\varphi_{e_H}=\text{id}_G$). Similarly

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$(g,h)(e_G,e_H)=\left(g\varphi_{h}(e_G),he_H\right)=(g,h)$

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(where we’ve used the fact that since $\varphi_h\in\text{Aut}(G)$ we have that $\varphi_h(e_G)=e_G$). Lastly, we need to prove that $\left(\varphi_{h^{-1}}(g^{-1}),h^{-1}\right)$ really is a two-sided inverse for every $(g,h)\in G\times H$. To do this merely perform the calculation

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\displaystyle \begin{aligned}(g,h)\left(\varphi_{h^{-1}}(g^{-1}),h^{-1}\right) &=\left(g\varphi_{h}\left(\varphi_{h^{-1}}(g^{-1}\right),hh^{-1}\right)\\ &=\left(g\varphi_{hh^{-1}}(g^{-1}),e_H\right)\\ &=\left(gg^{-1},e_H\right)\\ &=\left(e_G,e_H\right)\end{aligned}

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and similarly

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\begin{aligned}\left(\varphi_{h^{-1}}(g^{-1}),h^{-1}\right)(g,h) &= \left(\varphi_{h^{-1}}(g^{-1})\varphi_{h^{-1}}(g),h^{-1}h\right)\\ &= \left(\varphi_{h^{-1}}\left(g^{-1}g\right),e_H\right)\\ &= \left(e_G,e_H\right)\end{aligned}

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where we’ve used the fact that $\varphi_{h^{-1}}\in\text{Aut}(G)$.  The conclusion follows. $\blacksquare$

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Remark: Notice how amazingly unsymmetrical the calculations were, yet everything came out “right”. This is not an accident. The semidirect product is a weakening of the conditions of the direct product in an asymmetric way. The group $G$ is “nicer” in the above. This is reflected in the following theorem.

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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.

January 11, 2011 -