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Does there exist a surjective group homomorphism $\varphi:A_4\rightarrow\mathbb{Z}/4\mathbb{Z}$?

Edit: I have narrowed it down to the problem of whether $A_4$ has a normal subgroup of order 3 with 3-cycles as elements:

From the theorem of group homomorphism if $\varphi$ is a homomorphism then the $\operatorname{ker}(\varphi)$ has to be normal. From the isomorphism theorem the quotient group $A_{4} / \operatorname{ker}(\varphi)$ will be isomorphic to $\mathbb{Z} / 4 \mathbb{Z}$ (If the kernel is normal). Since $\mathbb{Z} / 4 \mathbb{Z}$ has order 4, we know that $A_{4} / \operatorname{ker}(\varphi)$ also has order 4 (since they're isomorphic). From lagranges theorem we can then conclude that $|\operatorname{ker} \phi|=\frac{\left|A_{n}\right|}{\left|A_{4} / \operatorname{ker} \phi\right|}=\frac{12}{4}=3$.
Since 3 is a prime and the order of a group element divides the order of the group, we know that the elements of the kernel must have order 3 (Not order 1, since all 1 cycles are the identity element). So the kernel has to be the group consisting of three 3-cycles from $A_4$.

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    $\begingroup$ Hint: what might be the kernel of such homomorphism? $\endgroup$
    – Mark
    May 19, 2022 at 21:48
  • $\begingroup$ @Mark I don't think I can say too much of what the kernel. Only that it has to be a normal subgroup of $A_4$ $\endgroup$
    – hh25
    May 19, 2022 at 22:17
  • $\begingroup$ @Mark Okay, I think I figured something out. From the theorem of group homomorphism then if $\varphi$ is homomorphism then the $ker(\varphi)$ has to be normal. From the isomorphism theorem then the quotient group $A_4/ker(\varphi)$ will be isomorphic to $\mathbb{Z} / 4 \mathbb{Z}$. Since $\mathbb{Z} / 4 \mathbb{Z}$ has order 4, we know that $A_4/ker(\varphi)$ also has order 4 (since they're isomorphic). From lagranges theorem we can then conclude that $|\operatorname{ker} \phi|=\frac{\left|A_{n}\right|}{\left|A_{4} / \operatorname{ker} \phi\right|}=\frac{12}{4}=3$. $\endgroup$
    – hh25
    May 19, 2022 at 22:51
  • $\begingroup$ How many elements does the kernel have? $\endgroup$
    – Berci
    May 19, 2022 at 22:51
  • $\begingroup$ since 3 is a prime and the order of a group element divides the order of the group we can say that the elements of the kernel must have order 3. So the kernel has to be the group consisting of 3 cycles in $A_4$ $\endgroup$
    – hh25
    May 19, 2022 at 22:53

1 Answer 1

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If $\varphi: G \to H$ is a group homomorphism, then given any element $a \in G$, we know that $\text{ord}(a)$ is a multiple of $\text{ord}(f(a))$.

Now we recall little background on the symmetric group $S_4$ and its subgroup $A_4$. The group $S_4$ has elements of cycles: 1-cycle, 2-cycle, 3-cycle, 4-cycles and product of two 2-cycles. Thus the possible orders of the elements of $S_4$ are $1,2,3,4$. But it is a fact that $A_4$ don't have an element of order $4$. So the possible orders of elements of $A_4$ are $1,2,3$.

If there is a surjective homomorphism $\varphi: A_4 \to \mathbb{Z}_4=\{\bar 0, \bar 1, \bar 2, \bar 3\}$. There exists some $a \in A_4$ such that $\varphi(a)=\bar 3$. Thus by the first sentence above, $\text{ord}(a)$ would be a multiple of $\text{ord}(\varphi(a))=\text{ord}(\bar 3)=4$, which is not possible, since $a \in A_4$ can not have order $4$. Thus there is no such surjective homomorphism.

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  • $\begingroup$ This is a nice way to show it. However, the initial fact that ord(f(a)) divides ord(a) for a group homomorphism is not something I have seen before (is not mentioned in my book or any exercises) and thus not something I would come up with. But I will try to see if I can prove this myself. $\endgroup$
    – hh25
    May 20, 2022 at 2:20
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    $\begingroup$ @hh25, thanks. It is a very basic property of group homomorphism. It should be found in any textbook covering group homomorphism. The proof is also simple. Look here $\endgroup$
    – MAS
    May 20, 2022 at 2:25

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