If $\phi$ is a homomorphism from $A_4$ onto a cyclic group $G,$ prove that $|G| = 1$ or $3.$

Question

Suppose $A_4$ as the set of 1-to-1 functions $\{1,2,3,4\}$ onto it self that the $sgn$ of them is even then the question states:

If $\phi$ is a homomorphism from $A_4$ onto a cyclic group G, prove that $|G| = 1$ or $3.$

We know that : $\ker\phi$ is a normal subgroup of $A_4$ and also from the first homomorphism theorem it is obvious that $A_4/\ker\phi$ should be cyclic so the length of $G$ can't be more than 3 ( because if $a\in A_4$ then $|a|\le3$ ) but I don't know how to show that $|G|\ne 2$

Can we say that there is no subgroup of 6 elements in $A_4$?

Answer 1

By the first isomorphism theorem, $G\cong A_4 / \ker \phi$.

We now check that the only normal subgroups of $A_4$ are of orders 1, 4, and 12 (the whole group). There are a few ways to do this (I'll list them in order of how elementary I think they are):

Union of Conjugacy Classes Method We know possible orders of subgroups are 1, 2, 3, 4, 6, 12 by Lagrange's theorem. Any normal subgroup is a union of conjugacy classes. The sizes of conjugacy classes in $A_4$ are 1, 3, 4 and 4.

(In particular, they are $\{(1)\},\{(1,2)(3,4),(1,3)(2,4),(1,4)(2,3)\},\{(1,2,3),(1,4,2),(1,3,4),(2,4,3)\}$, $\{(1,3,2),(1,2,4),(1,4,3),(2,3,4)\}$.)

It's not possible to add 1, 3, 4 and 4 to make 2, 3 or 6 using each number at most once (n.b. 1 must be used exactly once, since ${(1)}$ must be included in the union), so the only possible orders turn out to be 1, 4, and 12.

Sylow's Theorems Method for eliminating 3 If you know Sylow's theorems, 3 can be ruled out by the fact that any normal 3-subgroup would have to be a normal Sylow 3-subgroup, and hence the unique 3-subgroup. But there are at least two 3-subgroups, contradiction (for example, the subgroup generated by $(1,2,3)$ and the subgroup generated by $(2,3,4)$).

Homomorphism Method for Eliminating 6 We can also eliminate 6 as follows: suppose there was a normal subgroup $H$ of order 6. Then $A_4/H$ has order 2. Consider the projection homomorphism $\phi:A_4\to A_4/H$ given by $\phi(\sigma)=\sigma H$ for each $\sigma \in A_4$. Clearly $\ker \phi=H$ by construction. Let $\sigma$ be any 3-cycle. We claim $\phi(\sigma)=1$. Suppose not. Then $\phi(\sigma)$ must be the element of order 2 in $A_4/H$. But since $\phi$ is a homomorphism, the order of $\phi(\sigma)$ must divide the order of $\sigma$ i.e. 3. Therefore 2 divides 3, contradiction. We deduce that $\phi(\sigma)=1$ for every 3-cycle and hence every 3-cycle lies in $\ker \phi$. But as 3-cycles generate $A_4$, that must mean $\ker \phi=A_4$, contradicting the fact that $\ker \phi=H$.

Representation Theory Method If you know representation theory, the sizes of normal subgroups can just be read off the character table for $A_4$ (see e.g. https://people.maths.bris.ac.uk/~matyd/GroupNames/1/A4.html) - look at each row, and find all entries in the row which are equal to the leftmost entry. For each row, the union of the conjugacy classes corresponding to these entries gives a normal subgroup. We observe that they are of sizes 1, 4, and 12.

Now let's conclude the proof. Possible orders of $G\cong A_4 / \ker \phi$ are therefore 1, 3 and 12. We can rule out 12 as this would force $G$ to equal all of $A_4$ which is not cyclic. Therefore we are left with 1 and 3 as required.

Answer 2

With somewhat more sophistication and generality: let $\phi$ be a homomorphism from $A_4$ onto an abelian group $G$, then $|G|=1$ or $|G|=3$. Proof: put $N=ker(\phi)$. Then $G=A_4/N$ is abelian hence $[A_4,A_4] \subseteq N$. But the commutator subgroup $ [A_4,A_4] \cong V_4$, has order $4$. So, $|A_4:[A_4,A_4]|= 3$, and the result follows: $N=[A_4,A_4]$ or $N=A_4$.

Answer 3

To answer your specific question, note that there are $8$ $3$-cycles in $A_4$, which means there must exist some $3$-cycle $\sigma$ such that $\sigma \notin \ker \phi$. But because $[A_4: \ker \phi]=2$, that means $\sigma^2 \in \ker \phi$, which in turn means $(\sigma^2)^2 = \sigma^4 = \sigma \in \ker \phi$, which is a contradiction.

The answer of @jod is correct, but this allows you to reach the result without having to know that the $3$-cycles generate $A_4$.

Answer 4

Under a homomorphism, the order of the image of an element divides the order of the element. Therefore, if $|G|=2$, then the elements of order $3$ must lie in the kernel. But $A_4$ is generated by $3$-cycles, so also the elements of order $2$ lie in kernel, and hence $\phi$ is the trivial homomorphism: contradiction, because $\phi$ is onto $G$.

Answer 5

Can we say that there is no subgroup of 6 elements in $A_4$?

As you are dealing with the kernel, I assume you meant "no normal subgroup etc.". Sure: by this argument, the class equation of $A_4$ reads: $$12=1+3+4+4$$ and hence no chance to get a normal subgroup of order $6$$^\dagger$.

---

$^\dagger$Normal subgroups are union of conjugacy classes.