Help Understanding Explicit Automorphism Formulas for Semi-Direct Products 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?
 A: The setting was  $G \cong \mathbb Z/15\mathbb Z \rtimes_{\varphi} \mathbb Z/2\mathbb Z$, where $\varphi : \mathbb Z/2\mathbb Z \to \operatorname{Aut}(\mathbb Z/15\mathbb Z)$ is a group homomorphism.
1. Find an explicit description of $\varphi$
We have $$\operatorname{Aut}(\mathbb Z/15\mathbb Z) \cong (\mathbb Z/15\mathbb Z)^\times \cong (\mathbb Z/3\mathbb Z \times \mathbb Z/5\mathbb Z)^\times = (\mathbb Z/3\mathbb Z)^\times \times (\mathbb Z/5\mathbb Z)^\times \cong  \mathbb Z/2\mathbb Z\times \mathbb Z/4\mathbb Z,$$
such that we may (for the moment) think of $\varphi$ as a map $\mathbb Z/2\mathbb Z \to\mathbb Z/2\mathbb Z\times \mathbb Z/4\mathbb Z$
Clearly, $\varphi$ is determined by the image $\varphi(1)$.
Let us consider the example $\varphi(1) =(0,2) \in\mathbb Z/2\mathbb Z\times \mathbb Z/4\mathbb Z$.
To get the image $\varphi(1)\in \operatorname{Aut}(\mathbb Z/15\mathbb Z)$ in the original setting, we need to trace back the element $(0,2)\in\mathbb Z/2\mathbb Z\times \mathbb Z/4\mathbb Z$ in the above chain of isomorphisms.
First step: $(\mathbb Z/3\mathbb Z)^\times \times (\mathbb Z/5\mathbb Z)^\times \cong  \mathbb Z/2\mathbb Z\times \mathbb Z/4\mathbb Z$.
In $(0,2)\in\mathbb Z/2\mathbb Z \times \mathbb Z/4\mathbb Z$, the component $0$ is the neutral element in $\mathbb Z/2\mathbb Z$ and $2$ is the unique element of order $2$ in $\mathbb Z/4\mathbb Z$. Therefore, the preimage of the first component $0$ is the neutral element in $(\mathbb Z/3\mathbb Z)^\times$, which is $1$, and the preimage of the second component $1$ in $(\mathbb Z/5\mathbb Z)^\times$ is $-1$, which has order $2$. Together, the sought after preimage is $(1,-1)$.
Second step
$(\mathbb Z/15\mathbb Z)^\times \cong (\mathbb Z/3\mathbb Z)^\times \times (\mathbb Z/5\mathbb Z)^\times$
This is based on the Chinese remainder theorem $\mathbb Z/15\mathbb Z \cong \mathbb Z/3\mathbb Z\times \mathbb Z/5\mathbb Z$. For the preimage of $(1,-1)$ we need an integer having remainder $1$ mod $3$ and remainder $-1$ mod $5$. There is an algorithm for this computation (based on the Euclidean algorithm), but in this small case we simply check a few numbers and find
the preimage $4\in 15 \mathbb Z/\mathbb Z$.
Third step $\operatorname{Aut}(\mathbb Z/15\mathbb Z) \cong (\mathbb Z/15\mathbb Z)^\times$
This is easy. The concrete isomorphism (in the direction $\leftarrow$) is $a + \mathbb Z/15\mathbb Z \mapsto (x \mapsto ax)$. In our case, we get $x \mapsto 4x$
Fourth step
We have found $\varphi(1) = (x \mapsto 4x)$. What is $\varphi(a)$ for general $a\in\mathbb Z/2\mathbb Z$.
In this case, this is very easy, as the only other element of $\mathbb Z/2\mathbb Z$ is the neutral element $0$. Of cource, $\varphi(0) = \operatorname{id}_{\mathbb Z/15\mathbb Z}$.
For general $a\in\mathbb Z/2\mathbb Z$ we may combine this as as
$$\varphi(a) = (x \mapsto 4^a x).$$
(Remark. This would also work for larger cyclic groups instead of $\mathbb Z/2\mathbb Z$ using $a = 1 + \ldots + 1$ and the homomorphism property of $\varphi$.)
2. Determine the multiplication rule in the semidirect product
The general multiplication rule in the outer semidirect product $N \rtimes_{\varphi} U$ is
$$(n_1, u_1) \cdot (n_2, u_2) = (n_1\cdot \varphi(h_1)(n_2), h_1 \cdot h_2)$$
For elements $(a_1,b_1), (a_2,b_2)\in \mathbb Z/15\mathbb Z \times \mathbb Z/2\mathbb Z$ and $\varphi$ as above we get
$$(a_1,b_1) \cdot (a_2,b_2) = (a_1 + 4^{b_1} a_2,\, b_1 + b_2)$$
If you understand the above, you should be able to find the explicit multiplication rule also in the remaining cases.
3. Determine the group
We want to find out to which of the candidates $D_{15}$, $\mathbb Z/5\mathbb Z\times S_3$ and $\mathbb Z/3\mathbb Z\times D_5$ our group $\mathbb Z/15\mathbb Z\rtimes \mathbb Z/2\mathbb Z$ is isomorphic.
In general, checking two groups for isomorphism can be pretty hard.
But in our case, counting the elements of order $2$ should be sufficient to distinguish the groups.
We count the elements of $(a,b)\in\mathbb Z/15\mathbb Z\rtimes \mathbb Z/2\mathbb Z$ of order $2$. We consider the equation $(a,b)\cdot (a,b) = (0,0)$. The second component is $2b = 0$, which is always true. The first component is $a(1 + 4^b) = 0$. For $b = 0$ this reduces to $2a = 0$ with the only solution $a = 0$. For $b = 1$ we get $5a = 0$ with five solutions $a\in\{0,3,6,9,12\}$ possible. Discarding the neutral element $(0,0)$, we have found the following five elements of order 2:
$$\{(0,1), (3,1), (6,1), (9,1), (12,1)\}$$
Now you can count the elements of order $2$ in $D_{15}$, $\mathbb Z/5\mathbb Z\times S_3$ and $\mathbb Z/3\mathbb Z\times D_5$ to find out which one the considered semidirect product ist. (BTW, this also shows that the 3 candidate groups are pairwise non-isomorphic, as their number of elements of order $2$ differs.)
A: My answer is shorter.
If I understand you correctly, first we need to figure out what the automorphisms of the group $\mathbb{Z}_{15}$ look like.
Let $A=\langle5\rangle$, $B=\langle3\rangle$.  We have $\mathbb{Z}_{15}=A\oplus B$ and $|A|=3$, $|B|=5$.
Since $A$ and $B$ have coprime orders and $\operatorname{Aut}(A)\cong\mathbb{Z}_2$, $\operatorname{Aut}(B)\cong\mathbb{Z}_4$, it follows that
$$
\operatorname{Aut}(\mathbb{Z}_{15})
\cong\operatorname{Aut}(A)\times\operatorname{Aut}(B)
\cong\mathbb{Z}_2\times\mathbb{Z}_4.
$$
Therefore, it is natural to assume that
the automorphism $\alpha=(1,0)$ acts as follows
$\alpha(a)=-a$ for all $a\in A$ and $\alpha(b)=b$ for all $b\in B$ or
$$
\alpha(5q)=-5q,\ \alpha(3p)=3p,\ q=1,2;\ p=1,2,3,4.
$$
It can be given by a single formula $\alpha(1)=11$ (Check).
This means that $\mathbb{Z}_{15}\rtimes_{\varphi_1} K\cong D_3\times \mathbb{Z}_5$, where
$\varphi_1(t)=\alpha$ (we assume that $|K|=2$ and $K=\langle t\rangle$).
Similarly $\beta=(0,2)$ acts as follows
$\beta(a)=a$ for all $a\in A$ and $\beta(b)=-b$ for all $b\in B$ or
$$
\beta(5q)=5q,\ \beta(3p)=-3p,\ q=1,2;\ p=1,2,3,4;
$$
or $\beta(1)=4$ (Check).
If $\varphi_2(t)=\beta$, then $\mathbb{Z}_{15}\rtimes_{\varphi_2} K\cong \mathbb{Z}_3\times D_5$.
And the last
$\gamma=(1,2)$ acts as follows
$\gamma(a)=-a$ for all $a\in A$ and $\gamma(b)=-b$ for all $b\in B$
or $\gamma(1)=-1$.
If $\varphi_3(t)=\gamma$, then $\mathbb{Z}_{15}\rtimes_{\varphi_3} K\cong D_{15}$.
