Show that any nonabelian group $G$ of order $12$ which contains a normal subgroup of order $4$ must be isomorphic to $A_4$.

Hint. Show that $G$ is a split extension of a group of order $4$ by $\mathbb Z_3$, and, using the calculation of $Aut(\mathbb Z_2 \times \mathbb Z_2)$ given in Problem 14 of Exercises 4.6.13, show that here is only one such split extension which is nonabelian. (Exercise 4.6.13 tells us that $Aut(\mathbb Z_2 \times \mathbb Z_2) \cong S_3$.)

We know that $A_4 \cong (\mathbb Z_2 \times \mathbb Z_2) \rtimes_{\alpha} \mathbb Z_3$

Assume that $G$ is a group of order $12$ which has a normal subgroup $H$ of order $4$. Since $3\mid 12$ and is prime, we know that there exists a group of order $3$ in $G$. So we have an extension $f: \mathbb Z_3 \rightarrow G$ with kernel $H$. Since $H$ and $\mathbb Z_3$ are relatively prime, $f$ must be a split extension. Since split extensions are semidirect products, we just need to classify $H \rtimes_{\alpha} \mathbb Z_3$, right?

We know that $H \cong \mathbb Z_2 \times \mathbb Z_2$ or $\mathbb Z_4$.

Case 1:

$\mathbb Z_4 \rtimes_{\alpha} \mathbb Z_3$

We have $\alpha: \mathbb Z_3 \rightarrow Aut(\mathbb Z_4) \cong \mathbb Z_3$, and let $\mathbb Z_3 = \langle a \rangle$

$\bullet$ The homomorphism is trivial $\implies$ $\mathbb Z_4 \rtimes_{\alpha} \mathbb Z_3 \cong \mathbb Z_4 \times \mathbb Z_3$. Since both $\mathbb Z_4$ and $\mathbb Z_3$ are abelian, and the action is trivial; $\mathbb Z_3 \times \mathbb Z_3$ must also be abelian.

$\bullet$ $\alpha_1 : a \rightarrow a$

$\bullet$ $\alpha_2 : a \rightarrow a^2$

Let $\iota: H \rightarrow H \rtimes_{\alpha} K$, $\iota': H \rightarrow H \rtimes_{\alpha'} K$, $s: K \rightarrow H \rtimes_{\alpha} K$, and $s' : K \rightarrow H \rtimes_{\alpha'} K$

There is a proposition in our textbook that states:

Let $\alpha$ and $\alpha'$ be hoomorphisms of K into Aut(H). Then the following conditions are equivalent:

  1. $\alpha$ is conjugate to $\alpha' \circ g$, with $g \in Aut(K)$

  2. There is an isomorphism $\phi: H \rtimes_{\alpha} K \rightarrow H \rtimes_{\alpha'} K$ such that $im(\phi \circ \iota) = im \iota'$ and $im(\phi \circ s) = im s'$.

  3. there is a commutative diagram (I cannot draw this here but this is on page 116 here).

I'm not sure if I'm using 1. in the proposition correctly, but this is what I said about $\alpha1$ and $\alpha2$:

We know that $e\alpha_2e^{-1} = \alpha_1(a)$, so they must be conjugate.

Case 2:

We need to classify $(\mathbb Z_2 \times \mathbb Z_2) \rtimes_{\alpha} \mathbb Z_3$

We have $\alpha: \mathbb Z_3 \rightarrow Aut(\mathbb Z_2 \times \mathbb Z_2) \cong S_3$

$\bullet$ If $\alpha$ is trivial, we have $\mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_3 \cong \mathbb Z_2 \times \mathbb Z_6$. Since both $\mathbb Z_2$ and $\mathbb Z_6$ are abelian and the action is trivial, $\mathbb Z_2 \times \mathbb Z_6$ is also abelian.

We know that there are two elements in $S_3$ of order 3: (123) and (132).

So the other two homomorphisms must be:

$\bullet$ $\beta_1: a \rightarrow (123)$

$\bullet$ $\beta_2: a \rightarrow (132)$.

Now I want to show that $\beta_1$ and $\beta_2$ must be conjugate, but I'm stuck here.

My problem is that, in order to use the proposition, I need to have automorphisms, but since I'm using the fact that $S_3 \cong Aut(\mathbb Z_2 \times \mathbb Z_2)$, I'm not using automorphisms, I'm just using elements. So something like (for example) $\beta_1(13)$ makes no sense, right?

Basically, this was my plan:

I was going to show that $\beta_1$ and $\beta_2$ were conjugate. Since the problem is talking about nonabelian groups, the trivial homomorphisms of both cases don't really matter... since they both give abelian groups, right? So I would then try to prove that $(\mathbb Z_2 \times \mathbb Z_2) \rtimes_{\beta_1} Z_3 \cong \mathbb Z_4 \rtimes_{\alpha_1} \mathbb Z_3$, and then I would be done, right?

Thank you in advance

  • $\begingroup$ What is that text book, again...? $\endgroup$
    – DonAntonio
    May 22, 2013 at 23:18
  • $\begingroup$ @DonAntonio albany.edu/~mark/algebra.pdf $\endgroup$
    – user58289
    May 22, 2013 at 23:23
  • $\begingroup$ Z is a letter, while $\mathbb Z$ denotes the set of integers. $\endgroup$
    – user26857
    May 23, 2013 at 5:22

2 Answers 2


You did a pretty nice work but I think you have a fatal "little" mistake there: it must be $\;\text{Aut}(\Bbb Z_4)\cong\Bbb Z_2\;$ and not isomorphic to $\,\Bbb Z_3\,$ ...!

With this you're done since the only possible homomorphism $\;\Bbb Z_3\to\Bbb Z_2\,$ is the trivial one, which yields the abelian group $\,\mathbb Z_4\times \Bbb Z_3\,$...

Thus, a non-abelian group of order $\;12\;$ with a normal subgroup of order $\,4\,$ must be an extension of $\,\Bbb Z_2\times \Bbb Z_2\,$ and thus it must be $\,A_4\,$ .

  • $\begingroup$ Thank you so much. But don't I still need to show that $\beta_1$ and $\beta_2$ are conjugate? Because there must be only one nontrivial homomorphism (which is that of $A_4$), right? $\endgroup$
    – user58289
    May 23, 2013 at 8:28
  • 1
    $\begingroup$ Indeed, but I thought that was already known since you say "We know that $\,\Bbb (Z_2\times\Bbb Z_2)\rtimes \Bbb Z_3\,$ ...? Anyway, it never minds since in this case it is simpler, as you can read in page 49 in jmilne.org/math/CourseNotes/gt.html . You can see there that not only when two automorphisms are conjugate, but also when one is a shift (product) of the other by an autom. we get isomorphic semidirect products, and this is what you have here since $\,\beta_2=\beta_1^2\,$ ...! $\endgroup$
    – DonAntonio
    May 23, 2013 at 8:56

Here is an alternative solution not making use of the hint:

By the Sylow theorems, the number of $3$-Sylow subgroups of $G$ is either $1$ or $4$. The first case is not possible, since otherwiese $G$ would be abelian (as the direct product of the $3$-Sylow subgroup with the normal subgroup of order $4$).

Let $X$ be the set of the four $3$-Sylow subgroups. We look at the conjugation operation of $G$ on $X$, which induces an homomorphism $\varphi : G\to S_X$. By Sylow, the operation of $G$ is transitive. Let $P\in X$ be any $3$-Sylow subgroup. The orbit-stabilizer theorem yields $\left|N_G(P)\right| \cdot \left|X\right| = \left|G\right|$, which implies $\left|N_G(P)\right| = 3$ ($N_G(P)$ is the normalizer of $P$ in $G$). So $N_G(P) = P$. Since the order of $P$ is a prime, we get $\ker(\varphi) = \bigcap_{P\in X}N_G(P) = \{e\}$. Now the application of the homomorphism theorem to $\varphi$ yields $G\cong\varphi(G) \leq S_X \cong S_4$, so $G$ is isomorphic to an index $2$ subgroup of $S_4$. The only such subgroup is $A_4$.


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