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Suppose that $\phi$ is a homomorphism from $S_4$ onto $\mathbb Z_2$. Determine $\ker\phi$. Determine all homomorphisms from $S_4$ to $\mathbb Z_2$.

Please help me with this problem. I am stuck and I feel like I am not making any progress...

Also, unlike isomorphism, is it okay for one group to be non-cyclic and the other to be cyclic under homomorphisms? Because I think $S_4$ is not cyclic but $Z_2$ is cyclic.

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marked as duplicate by Claude Leibovici, Joey Zou, user26857 abstract-algebra Dec 1 '16 at 9:18

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  • $\begingroup$ Hint. The kernel of such a homomorphism must be a normal subgroup of $S_4$. How many cosets must it have? What are the possibilities? $\endgroup$ – Ethan Bolker Nov 30 '16 at 21:42
  • $\begingroup$ @EthanBolker I still can't figure it out using the fact that Kernel is a normal subgroup. But aren't permutations of order 3 must be in the Kernel? Since $\phi (a^3) = \phi (a) + \phi (a) + \phi (a) = 0$, and this is only possible when $\phi (a) = 0$. Although, other permutations of different order could be in the Kernel as well. Since for permutations of order 2, $\phi(a^2) = \phi(a) + \phi(a) = 0$ is possible either when $\phi(a) = 1$or $0$ $\endgroup$ – user3000482 Dec 1 '16 at 4:15
  • $\begingroup$ @AlanWang I already looked at that, but it doesn't ask how to find the Kernel. $\endgroup$ – user3000482 Dec 1 '16 at 4:18
  • $\begingroup$ In the second answer given in that question, $|\ker\phi|=12$. Try to find possible normal subgroup of $S_4$ with order $12$. $\endgroup$ – Alan Wang Dec 1 '16 at 4:23
  • $\begingroup$ @AlanWang Could you please give me a hint? I really have no clue to find the normal subgroups of $S_4$, I don't even know how to find the subgroups of $S_4$. I only know $D_4$ is a subgroup. $\endgroup$ – user3000482 Dec 1 '16 at 4:24
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If $\phi:S_4 \to \Bbb Z_2$ is a homomorphism, its image must be a subgroup of $\Bbb Z_2$. Since $\Bbb Z_2$ only has two subgroups, itself, and the trivial subgroup $\{0\}$, this simplifies things enormously.

If the image is the trivial group, then $\phi$ is the $0$-homomorphism that maps everything to the identity of $\Bbb Z_2$. So that's one possibility.

Otherwise, by the first isomorphism theorem, we have:

$\Bbb Z_2 =\phi(S_4) \cong S_4/\text{ker }\phi$.

This means that $2 = |\Bbb Z_2| = |S_4|/|\text{ker }\phi| = 24/|\text{ker }\phi|,$

That is: $|\text{ker }\phi| = 24/2 = 12$. So the kernel (if there is such a non-trivial homomorphism $\phi$) is a subgroup of $S_4$ of order $12$.

Now, if $\sigma \in S_4$ is a $3$-cycle, then $0 = \phi(e) = \phi(\sigma^3) = (\phi(\sigma))^3$, that is the order of $\phi(\sigma)$ divides $3$, so has order $1$ or $3$, by a corollary to Lagrange's theorem. We can rule out an order of $3$ since $3\not\mid 2$, so every $3$-cycle is in the kernel of $\phi$.

Counting, we see we have $8\ 3$-cycles in $S_4$, so those (plus the identity) account for $9$ elements of our $12$-element kernel (should it exist).

Noting that $(a\ b\ c)(a\ b\ d) = (a\ c)(b\ d)$, we can see that (by properly choosing $a,b,c,d$) every $2,2$-cycle is also in the kernel of $\phi$:

$\phi((a\ c)(b\ d)) = \phi((a\ b\ c)(a\ b\ d)) = \phi((a\ b\ c)) + \phi((a\ b\ d)) = 0 + 0 = 0$.

Since there are three of these, we have found our $12$-element kernel, provided that:

$\{e, (1\ 2\ 3), (1\ 3\ 2), (1\ 2\ 4), (1\ 4\ 2), (1\ 3\ 4), (1\ 4\ 3), (2\ 3\ 4), (2\ 4\ 3), (1\ 2)(3\ 4), (1\ 3)(2\ 4), (1\ 4)(2\ 3)\}$

actually forms a subgroup of $S_4$, which it does, this is the so-called alternating subgroup $A_4$ of all even permutations on the set $\{1,2,3,4\}$.

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If $G$ and $H$ are groups and $\varphi : G \to H$ is a homomorphism, then for any $x \in G$ the order of $\varphi(x)$ in $H$ must divide the order of $x$ in $G$. Thus if $\varphi : S_4 \to \mathbb{Z}_2$ is a homomorphism, $\ker \varphi$ must contain all of the $3$ cycles. Do you know what group the $3$ cycles generate?

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  • $\begingroup$ I can see that 3 cycles must be in the ker $\varphi$, but aren't 2-cycles and 4-cycles possible to be in the kernel as well? Since for permutations of order 2, $\varphi(a^2) = \varphi(a) + \varphi(a) = 0$ and this is possible either when $\varphi(a)$ = 0 or 1. $\endgroup$ – user3000482 Dec 1 '16 at 4:22
  • $\begingroup$ Correct. So all the $3$ cycles are in $\ker\varphi$. Now suppose $a$ is a $2$-cycle. As you pointed out, $\ker\varphi(a)$ can be either $0$ or $1$. I claim that once you fix it, the rest of $\varphi$ is determined. So there are only $2$ possible homomorphisms. $\endgroup$ – Ethan Alwaise Dec 1 '16 at 4:24
  • $\begingroup$ Can same thing be said for permutations of order 4? I am not sure what you mean by fixing it... could you please elaborate on your statement little more? $\endgroup$ – user3000482 Dec 1 '16 at 4:37
  • $\begingroup$ I am saying that once you know the value of $\varphi$ at any even permutation, you know the whole homomorphism. Try figuring out the size of the group generated by the $3$-cycles. That will help a lot. $\endgroup$ – Ethan Alwaise Dec 1 '16 at 5:10

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