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I was reading the classification of groups of order 30 from Dummit & Foote(pg-182, 3rd ed). It has a normal subgroup of order 15 which is obviouly cyclic. Hence G is isomorphic to $\mathbb{Z}_{15} \rtimes_{\phi} \mathbb{Z}_2$ by recognition theorem of semi direct products where $\phi :\mathbb{Z}_2 \rightarrow Aut(\mathbb{Z}_{15})$ is a homomorphism. Since Aut$(\mathbb{Z}_{15})$ is isomorphic to $\mathbb{Z}_4 \times \mathbb{Z}_2$ we see we have three distinct homomorphisms $\phi$ call them $\phi_1$ ,$\phi_2$, $\phi_3$. Now let a be a generator of $\mathbb{Z}_5$ and b is the generator of $\mathbb{Z}_3$ and let k be the generator of $\mathbb{Z}_2$ and define $\phi_1 :\mathbb{Z}_2 \rightarrow Aut(\mathbb{Z}_{15})$ such that k maps to the following automorphism of $\mathbb{Z}_5 \times \mathbb{Z}_3$

$\mathbb{Z}_5 \times \mathbb{Z}_3$ $\rightarrow$ $\mathbb{Z}_5 \times \mathbb{Z}_3$

a $\rightarrow$ a

b $\rightarrow$ $b^{-1}$

My question is that why the semidirect product $\mathbb{Z}_{15} \rtimes_{\phi_1} \mathbb{Z}_2$ is isomorphic to $\mathbb{Z}_5 \times D_6$ . I dont know how to show this isomorphism explicitly. I would like an elaborate explanation for this , that wil make the concepts clear. Thanks in advance!

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We can write all elements of $(\mathbb Z_5 \times \mathbb Z_3) \rtimes_{\phi_1} \mathbb Z_2$ as triples $(a,b,c)$ where $a \in \mathbb Z_5,$ $b \in \mathbb Z_3,$ and $c \in \mathbb Z_2.$

The definition of semidirect product then says that if we want to multiply two of these elements, we get \begin{align} (a_1, b_1, c_1) \cdot (a_2, b_2, c_2) &= ((a_1, b_1) \cdot \phi_1(c_2)(a_2, b_2), c_1 c_2) \\ &= ((a_1, b_1) \cdot (a_2, \phi_*(c_1)(b_2)), c_1 c_2) \\ &= (a_1 a_2, b_1 \phi_*(c_1)(b_2), c_1 c_2). \end{align} where by $\phi_*$ I mean the map $\mathbb Z_2 \to \operatorname{Aut}(\mathbb Z_3)$ which sends the non-identity element of $\mathbb Z_2$ to the taking-the-inverse automorphism of $\mathbb Z_3$.

So we see that the $a_1$ and $a_2$ coordinates don't interact at all with the $(b_1,c_1)$ and $(b_2,c_2)$ coordinates, and so this is a direct product in which the first factor is $\mathbb Z_5$.

The second factor is a semidirect product in which $$(b_1, c_1) \cdot (b_2, c_2) = (b_1 \phi_*(c_1)(b_2), c_1, c_2),$$ so in other words, it is $\mathbb Z_3 \rtimes_{\phi_*} \mathbb Z_2$. But this is precisely the definition (a definition, anyway) of $D_6$.

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