Suppose \(J\) and \(L\) are groups, and \(\varphi: L \to \mathrm{Aut}(J)\) is a group homomorphism.
Give a definition for the group \(J \rtimes_\varphi L\).
The identity function gives a bijection of sets \(J \rtimes_\varphi L \to J \times L\). Prove that this function is a group isomorphism if and only if \(\varphi\) is the trivial homomorphism sending everything to the identity.
Solution
Write \(\varphi_\ell=\varphi(\ell)\in\operatorname{Aut}(J)\) for \(\ell\in L\).
1. The semidirect product is the set \(J\times L\) with multiplication\[ (j,\ell)*(j',\ell')= \bigl(j\varphi_\ell(j'),\ell\ell'\bigr). \]
For three pairs, multiplying the first two and then the third gives \[ \bigl(j\varphi_\ell(j')\varphi_{\ell\ell'}(j''), \ell\ell'\ell''\bigr). \] Multiplying the last two and then the first gives \[ \bigl(j\varphi_\ell(j'\varphi_{\ell'}(j'')), \ell\ell'\ell''\bigr). \] These are equal because \(\varphi_\ell\) preserves multiplication and \(\varphi_{\ell\ell'}=\varphi_\ell\circ\varphi_{\ell'}\). Thus the operation is associative.
The identity is \((1_J,1_L)\), since \(\varphi_{1_L}=\operatorname{id}_J\) and each automorphism fixes \(1_J\). The inverse of \((j,\ell)\) is \[ \bigl(\varphi_{\ell^{-1}}(j^{-1}),\ell^{-1}\bigr). \] Multiplication in the first order gives first coordinate \(j\varphi_\ell\varphi_{\ell^{-1}}(j^{-1})=1_J\); in the reverse order it gives \(\varphi_{\ell^{-1}}(j^{-1})\varphi_{\ell^{-1}}(j)=1_J\). Both second coordinates are \(1_L\). This verifies all group axioms.
2. The identity function on \(J\times L\) is a group isomorphismfrom this semidirect product to the direct product if and only if\(\varphi\) is trivial.
If every \(\varphi_\ell\) is the identity on \(J\), the multiplication in step 1 becomes \((j,\ell)*(j',\ell')=(jj',\ell\ell')\), the direct-product multiplication. The identity function is then an isomorphism.
Conversely, suppose this identity function preserves multiplication. For any \(j\in J\) and \(\ell\in L\), multiplication in the semidirect product gives \[ (1_J,\ell)*(j,1_L)=(\varphi_\ell(j),\ell), \] whereas multiplication of the same pairs in the direct product gives \((j,\ell)\). Preservation by the identity map implies \(\varphi_\ell(j)=j\) for every \(j,\ell\). Thus every \(\varphi_\ell\) is the identity automorphism, proving the converse.