I'm currently working through some material on semi-direct products, and I'm totally confused by some of the explicit examples given for semi-direct products, namely how one determines the automorphisms.
For example, In classifying groups of order $30$, done here specifically, the answer makes sense in that there are four possible automorphisms given by $\varphi(1) \in \{(0,0), (1,0), (0,2), (1,2)\}$, but what are these automorphisms?
For $\varphi(1) = (0,0)$, I get this is a trivial map corresponding to simply the direct product.
What about $\varphi(1) = (1,0)$, what's an explicit formula for this automorphism? Semi-direct products of this form arise from the action of conjugation of the right factor $K$ on the left factor $H$ (for a semi direct product $H \rtimes_{\varphi} K$). In this case, we have a Sylow $2$ subgroup $P_2 \cong \mathbb{Z}_2$, and a group $N \cong \mathbb{Z}_{15}$ of order $15$.
The theory states that if $\varphi: K \longrightarrow \text{Aut}(H)$, then $\varphi(k)[(h)] = khk^{-1}$. So if $\varphi(1) = (1,0)$, then wouldn't we simply have for $n \in \mathbb{Z}_{15}$: $$\varphi(1)(n) = 1+n-1 = n?$$ This is the trivial automorphism however, which is in direct contradiction to the fact that $\varphi(1) = (1,0)$ isn't trivial.
Essentially what I'm asking for is an explanation on how one deduces that (in the context of the linked post): "if $\varphi(1)=(1,0)$, then
$P_2 \rtimes_{\varphi} N \cong D_{15}$ because _____, and if $\varphi(1)=(0,2)$, then $P_2 \rtimes_{\varphi} N \cong \mathbb{Z}_5 \times S_3$ because ___…"
Could someone explain what's going on here?
Best Answer
Groundwork
We are trying to understand semidirect products of the form $ℤ_{15} ⋊_φ ℤ_2$. We rely on the following observations:
Let $R$ be a ring and let $u$ be an element of $R^×$. The map $$ λ_u \colon R \to R \,, \quad x \mapsto ux $$ is an automorphism of the underlying additive group of $R$. We have thus a homomorphism of groups $$ λ \colon R^× \to \operatorname{Aut}_ℤ(R) \,, \quad u \mapsto λ_u \,. $$ This homomorphism is always injective because $u$ can be retrieved from $λ_u$ as $u = λ_u(1)$. If $R = ℤ_n$, then $λ$ is also surjectiv, and thus an isomorphism of groups $\operatorname{Aut}_ℤ(ℤ_n) ≅ ℤ_n^×$.
Given two coprime integers $m$ and $n$, we have an isomorphism of rings (and thus in particular of abelian gorups) $ℤ_{mn} ≅ ℤ_m × ℤ_n$ given by $[k] \mapsto ([k], [k])$.
For every two rings $R$ and $S$, we have $(R × S)^× = R^× × S^×$.
Given any group $G$, homomorphism of groups from $ℤ_2$ is $G$ are uniquely determined by their image of the element $1$ of $ℤ_2$. This image can be any element of $G$ whose order divides $2$. (In other words, this element can either be the neutral element of $G$, or an element of order $2$. The neutral element corresponds to the trivial homomorphism from $ℤ_2$ to $G$.)
Let us fix a piece of notation:
We will need one very explicit example of this general construction:
We will also need the following observation:
Computation of elements
Let us examine the chain of isomorphisms $$ ℤ_{15}^× ≅ (ℤ_3 × ℤ_5)^× = ℤ_3^× × ℤ_5^× ≅ ℤ_2 × ℤ_4 $$ in more detail:
The isomorphism $ℤ_{15} ≅ ℤ_3 × ℤ_5$ comes from the chinese reminder theorem because $3$ and $5$ are coprime. More explicitly, the map $$ θ \colon ℤ_{15} \to ℤ_3 × ℤ_5 \,, \quad [n] \mapsto ([n], [n]) $$ is an isomorphism. We can also give an explicit formula for the inveres of $θ$: we have $θ^{-1}(1, 0) = 10 = -5$ and $θ^{-1}(0, 1) = 6$, and therefore $$ θ^{-1}([m], [n]) = [-5m + 6n] \,. $$
The groups $ℤ_3^×$ and $ℤ_5^×$ are cyclic, and generated by the elements $2$ and $2$ respectively. We can therefore choose the isomorphism $ℤ_3^× × ℤ_5^× ≅ ℤ_2 × ℤ_4$ so that an element $([k], [l])$ of $ℤ_2 × ℤ_4$ corresponds to the element $(2^k, 2^l)$ of $ℤ_3^× × ℤ_5^×$.
For an element $([k], [l])$ of $ℤ_2 × ℤ_4$ we therefore find that
We have now the following elements in the groups $ℤ_2 × ℤ_4$, $ℤ_3^× × ℤ_5^×$ and $ℤ_{15}^×$ whose order divides $2$.
For every element of $ℤ_2 × ℤ_4$, the corresponding automorphisms of $ℤ_{15}$ are simply given by multiplication with corresponding element of $ℤ_{15}^×$. As an example, the automorphism of $ℤ_{15}$ corresponding to the element $(1, 0)$ of $ℤ_4 × ℤ_2$ is given by multiplication with $-4$.
Computation of semidirect products
Let us now compute the resulting semidirect products.
The element $(0, 0)$ of $ℤ_2 × ℤ_4$ corresponds to the element $1$ of $ℤ_{15}^×$. The resulting automorphism of $ℤ_{15}$ is simply the identity automorphism, and the resulting homomorphism from $ℤ_2$ to $\operatorname{Aut}(ℤ_{15})$ is the trivial homomorphism. And the semidirect product resulting from the trivial homomorphism is simpli the direct product. Therefore, $$ ℤ_{15} ⋊_1 ℤ_2 ≅ ℤ_{15} × ℤ_2 ≅ ℤ_{30} \,. $$
The element $(0, 2)$ of $ℤ_2 × ℤ_4$ corresponds to the element $4$ of $ℤ_{15}$. To compute the resulting semidirect product $ℤ_{15} ⋊_4 ℤ_2$, we employ once again the isomorphism $ℤ_{15} ≅ ℤ_3 × ℤ_5$. The corresponding element in $ℤ_3 × ℤ_5$ is given by $(1, -1)$. We have therefore an isomorphism of groups $$ ℤ_{15} ⋊_4 ℤ_2 ≅ (ℤ_3 × ℤ_5) ⋊_{(1, -1)} ℤ_2 ≅ ℤ_3 × (ℤ_5⋊_{-1} ℤ_2) ≅ ℤ_3 × \mathrm{D}_5 \,. $$
For the element $(1, 0)$ of $ℤ_2 × ℤ_4$ we find similarly that $$ ℤ_{15} ⋊_{-4} ℤ_2 ≅ (ℤ_3 × ℤ_5) ⋊_{(-1, 1)} ℤ_2 ≅ (ℤ_3 ⋊_{-1} ℤ_2) × ℤ_5 ≅ \mathrm{D}_3 × ℤ_5 ≅ \mathrm{S}_3 × ℤ_5 \,. $$
For the element $(1, 2)$ of $ℤ_2 × ℤ_4$ the corresponding element of $ℤ_{15}^×$ is given by $-1$. We find that $$ ℤ_{15} ⋊_{-1} ℤ_2 ≅ \mathrm{D}_{15} \,. $$