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I was trying to classify all groups of the form $\mathbb Z_{21} \rtimes_{\alpha} \mathbb Z_2$ and show that these groups are $\mathbb Z_{42}$, $D_{42}$, $D_6\times \mathbb Z_7$, and $\mathbb Z_3\times D_{14}$.

I said that if we have $\mathbb Z_{21} \rtimes_{\alpha} \mathbb Z_2$, then we have a homomorphism $\alpha: \mathbb Z_2 \rightarrow \operatorname{Aut}(Z_{21})$. But $\operatorname{Aut}(\mathbb Z_{21}) \cong \mathbb Z^{\times}_{21}$. But we know that $\mathbb Z^{\times}_{21}$ is either isomorphic to $\mathbb Z_{12}$ or $\mathbb Z_2\times \mathbb Z_6$.

Case 1:

We look at $\mathbb Z_2\times \mathbb Z_6$, and so we have the homomorphism $\alpha: \mathbb Z_2 \rightarrow \mathbb Z_2\times \mathbb Z_6$. Let $\mathbb Z_2 = <a>$, $\mathbb Z_2=<a>$, and $\mathbb Z_6=<b>$.

So the possibilities for the homomorphisms are:

a) $\alpha(a) = (e,e)$

b) $\alpha(a)= (a,e)$

c) $\alpha(a) = (a,b^3)$

But know I'm stuck. I'm not sure how I would classify groups from these homomorphisms. In other words, how do I know which homomorphism corresponds to which group? I looked at some examples in the internet, but they did not show the step that I'm looking for. They just gave the possible homomorphisms and then skiped that step. So I'm not sure how I can approach this.

Thanks in advance

I don't know why the part that I wrote for the bounty does not have any spaces, so I decided to copy and paste it here so it is easier to read

1) I'm trying to connect Caranti's answer with Easy's. I understand Easy's answer, because for example, if $\alpha$ acts trivially on the first coordinate, then $(Z_3 \times Z_7) \rtimes Z_2 = (Z_3 \rtimes Z_2) \times (Z_7 \rtimes Z_2) = (Z_3 \rtimes Z_2) \times Z_7$. And we know that $Z_n \rtimes Z_2$ is $D_{2n}$ for any n, right? So we have $Z_3 \rtimes $Z_2 = D_6$. Now I'm trying to relate this to the technique that Caranti used, but I'm not sure how. In other words, I don't understand the step from where he found the automorphisms to the step where he got the corresponding groups.

2) We can also think about this question in terms of split extensions, right? So we have the split extension $\Bbb{Z}_{21} \rightarrow G \rightarrow G/\Bbb{Z}_{21}$ where G is of order 42. Now let $f: G \rightarrow G/\Bbb{Z}_{21}$. Let $H = \Bbb{Z}_{21}$ and $K = \Bbb{Z}_2$ We know that G/H is of order 2. So if we have a section $s: G/H \rightarrow G$, for the nontrivial element $a \in G/H$, are supposed to send a to an element in G of order 2, right?

But there is something that's confusing me about this. In the textbook, it says:

Let $ε : S_n → {±1}$ take each permutation to its sign. Then ε is an extension of $A_n$ by $Z_2$. Moreover, setting s(−1) = τ ∈ $ S_n$ determines a section of ε if and only if τ has order 2 and is an odd permutation. Since the order of a product of disjoint cycles is the least common multiple of the orders of the individual cycles, we see that τ determines a section of ε if and only if it is the product of an odd number of disjoint 2-cycles. For such a τ , we write $s_τ$ for the section of ε with $s_τ (−1) = τ$ . Given the variety of such τ, we see that there are many possible choices of section for ε…if n ≥ 4, then each different choice for a section of ε induces a different homomorphism from $Z_2$ to $Aut(A_n)$.

Thus, the classification of even the split extensions of a nonabelian group offers substantial difficulty.

Lemma 4.7.9. Let f : G → K be an extension of H by K, where H is abelian. For x ∈ G, write $c_x$ : H → H for the automorphism obtained from conjugation by x. Suppose that f(x) = f(y). Then $c_x = c_y$ as automorphisms of H. Moreover, there is a homomorphism $α_f : K → Aut(H)$ defined by setting $α_f (k)$ equal to $c_x$ for any choice of x with f(x) = k. In particular, if f happens to be a split extension and if s : K → G is a section of f, then the homomorphism from K to Aut(H) induced by s coincides with the homomorphism $α_f$ above. Thus, if H is abelian, the homomorphism from K to Aut(H) induced by a section of f is independent of the choice of section.

We can see that H is abelian in this case, so why do we need to consider every single case of where a is sent? In other words, if we need to consider every case of where a can be sent, then what is the difference between this problem (where H is abelian) from the problem given in the example where H is not abelian?

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up vote 6 down vote accepted

Look, $\Bbb{Z}_{21}^{\times}$ is isomorphic to $\Bbb{Z}_{2} \times \Bbb{Z}_{6}$ by CRT, so there's only this case to consider.

Now let us identify the elements of order dividing $2$ in $\Bbb{Z}_{21}^{\times}$. These are (the classes of) $$ 1, -1, 8, -8. $$ This you can see by solving the equation $$x^{2} \equiv 1 \pmod{21}.\tag{sqrt}$$ This is easy enough in this case to be done by inspection, but the fact that there are four roots in this case, and a general method to solve $x^{2} \equiv m \pmod{n}$ when a factorization of $n$ is available, comes from the CRT.

So the corresponding automorphisms of $\Bbb{Z}_{21}$ are $$ \text{the identity}, \quad a : n \mapsto -n, \quad b : n \mapsto 8 n, \quad ab : n \mapsto -8n. $$ They correspond to your four groups, in the order $$ \Bbb{Z}_{21} \times \Bbb{Z}_{2} \cong \Bbb{Z}_{42}, \quad D_{42}, \quad \Bbb{Z}_3\times D_{14}, \quad D_6\times \Bbb{Z}_7. $$ This can be seen by noting that the $\Bbb{Z}_3$ within the $\Bbb{Z}_{21}$ is generated by (the class of) $7$, and the $\Bbb{Z}_7$ by $3$.

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Thanks, but how do you know that you have to look at elements that divide 2? Is that related to the inverse homomorphism that Easy mentioned? – user58289 Apr 5 '13 at 11:43
@Artus, you are looking at all possible homomorphisms $\alpha: \Bbb{Z}_2 \rightarrow \operatorname{Aut}(\Bbb{Z}_{21})$. Now such an $\alpha$ maps the generator of $\Bbb{Z}_2$ (which is an element of order $2$) into an element of order $1$ or $2$. – Andreas Caranti Apr 5 '13 at 11:51
@Artus, yes, these four elements are the possible images of the generator of $\Bbb{Z}_{2}$ under $\alpha$, so they are automorphisms of $\Bbb{Z}_{21}$, functions then. The first one is the identity, the second one the map that takes $n \in \Bbb{Z}_{21}$ to $-n$ and so on. The notation I use is common for maps, by writing $b : n \mapsto 8 n$ I mean that $b(n) = 8n$ for instance. – Andreas Caranti Apr 5 '13 at 15:20
@Artus, I will try and answer when my time zone permits. – Andreas Caranti May 22 '13 at 21:41
@Artus, first of all $\Bbb{Z}_{21}$ is an additive group, so inversion is $f(x) = -x = (-1) \cdot x$. Then, $\Bbb{Z}_{21}$ is cyclic, generated by $1$, so any endomorphism $\alpha$ is determined by the image $a$ of $1$, which can be taken arbitrarily. Now if $\alpha(1) = a$, then $\alpha(x) = x \alpha(1) = a x$. Such an endomorphism is invertible (that is, it is an automorphism) if and only if $a$ is invertible in $\Bbb{Z}_{21}$, that is, $a \in \Bbb{Z}_{21}^{\times}$. – Andreas Caranti May 23 '13 at 6:28

Consider that $\text{Aut}(\mathbb{Z}_{21})=\text{Aut}(\mathbb{Z}_3\times\mathbb{Z}_7)=\text{Aut}(\mathbb{Z}_3)\times\text{Aut}(\mathbb{Z}_7)=\mathbb{Z}_2\times\mathbb{Z}_6$. Write the elements in $\mathbb{Z}_{21}$ as $(x,y)$ where $x\in\mathbb{Z}_3,y\in\mathbb{Z}_7$, then $\alpha\in\mathbb{Z}_2$ acts on the first or the second coordinate or both of them.

  1. If $\alpha$ acts trivially on both coordinates, then $\alpha$ centralises $\mathbb{Z}_{21}$, and so $G=\mathbb{Z}_{21}\times\mathbb{Z}_2=\mathbb{Z}_{42}$.

  2. If $\alpha$ acts only on the first coordinate, then $G=(\mathbb{Z}_3\rtimes\mathbb{Z}_2)\times\mathbb{Z}_7=D_6\times\mathbb{Z}_7$.

  3. If $\alpha$ acts on the second, then $G=\mathbb{Z}_3\times D_{14}$.

  4. If $\alpha$ acts on both, then $G=D_{42}$.

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Thanks a lot. I have two more questions (if you don't mind). First, why do we only look at inverse mapping? I thought we were supposed to look at all possible homomorphisms when classifying these. Second, is there a way to know that $Z_3 \rtimes Z_2 \cong D_6$ by inspection. Or is it someting I have to prove? – user58289 Apr 5 '13 at 11:38
@Artus, I think I actually don't need inverse mapping here, so I just delete it. For $\mathbb{Z}_3\rtimes\mathbb{Z}_2$, since $0\in\mathbb{Z}_3$ is fixed by any automorphism, so the only automorphism of it is just swapping the two nonzero elements, and hence unique. In fact, any automorphism of a cyclic group is just a scalar multiplication, and will be determined by the order, so if you find one, then it should be the one. – Easy Apr 5 '13 at 11:44

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