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Find a homomorphism $phi$ from $U(40)$ to $U(40)$ with kernel ${1,9,17,33}$ and $phi(11)=11$

Iam looking for a solution since I have already tried several times myself.

RFZ
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Malcolm
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  • Hint: If the homomorphism is not trivial, how large must the kernel be? – Tobias Kildetoft Apr 24 '14 at 17:43
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    What defines a trivial homomorphism? @TobiasKildetoft, im not quite sure how large the kernel can be, what am i missing? – Malcolm Apr 24 '14 at 17:52
  • By trivial I mean the one that sends everything to the neutral element. For finding the size of the kernel, use an isomorphism theorem (possibly the first, but the numbering is not consistent). – Tobias Kildetoft Apr 24 '14 at 17:53

1 Answers1

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The key to figuring this out is to remember that knowing all homomorphic images of a group is equivalent to knowing all the normal subgroups of a given group (think about the kernel). The theorem that should be familiar to you is

Theorem: Let $ \phi : \mathcal{G} \rightarrow \bar{\mathcal{G}} $ be a homomorphism with kernel $K$. Then $ \mathcal{G} / K \approx \bar{\mathcal{G}} $.

What the above says is exactly what I said originally, namely knowing all the normal subgroups of a group gives one all the homomorphic images of a group. This is because a normal subgroup can be exclusively defined as the kernel of a homomorphism.

In your case find all normal subgroups of $S_4$ and proceed from there.

o0BlueBeast0o
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  • Since we are just looking for $S_4\to\mathbb Z_2$, the images will be of order $1$ or $2$, so the kernels… – Christoph Apr 24 '14 at 18:55
  • If you are familiar with group homomorphisms then you should be familiar with a theorem which states: If $ \phi : \mathcal{G} \rightarrow \bar{\mathcal{G}} $ with kernel $K$, then $ \mathcal{G} / K \approx \bar{\mathcal{G}}$. Since a normal subgroup is defined as the kernel of a group homomorphism, knowing all the normal subgroups then looking at the quotient groups gives one all homomorphic images of a group – o0BlueBeast0o Apr 24 '14 at 23:07
  • What I was hinting at is that, since the only divisors of $2$ are $1$ and $2$, we only need to find normal subgroups of order $24$ (trivial) and $12$, not all the normal subgroups. – Christoph Apr 25 '14 at 05:49
  • @o0BlueBeast0o, im thinking that if $\phi$ is a homomorphism then $ker(\phi)$ must be a normal subgroup of $S_4$, since only 2 such normal subgroups exist we have found our two homomorphisms. Is this correct? – Malcolm Apr 25 '14 at 19:24
  • Essentially yes. What the theorem states is that finding the normal subgroups of $S_4$ and then looking at the corresponding quotient groups, $S_4 / N $, will give you the homomorphic images of the group. Of course there is the homomorphism which maps everything to the identity, this corresponds to $S_4 / S_4 \approx \left< e \right>$. Now we look for normal subgroups which gives a quotient group that looks like $\mathbb{Z}_2$. We know that the alternating group $A_n$ is always normal in $S_n$ and has order $\frac{1}{2} n!$. $S_n / A_n$ will be the quotient group isomorphic to $\mathbb{Z}_2$ – o0BlueBeast0o Apr 25 '14 at 21:59
  • @o0BlueBeast0o, thank you very much. But how can i argue when finding the normal subgroups of $S_4$? Of course I know them (only 2 exist?), but I have to give an argument of how i found them? – Malcolm Apr 25 '14 at 22:05
  • Well my answer really depends on how much groups theory you know. There are two non-trivial normal subgroups yes. If you've developed permutation groups it is rather easy to argue why $A_n$ is normal. This is because $A_n$ consists of all even permutations in $S_n$. Conjugation will not change the parity of a even permutation. This says that $A_n$ is normal in $S_n$. The other non-trivial normal subgroup (which is a four group I believe) shouldn't play a role since its quotient group will have order 6, which is obviously to large for $\mathbb{Z}_2$ – o0BlueBeast0o Apr 25 '14 at 22:10
  • @o0BlueBeast0o according to my calculations I have found that ${(),(1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}$ and $A_4$ aswell as just ${e}$ are the only normal subgroups of $S_4$, is this correct? – Malcolm Apr 26 '14 at 00:00
  • Yes also remembering that $S_4$ is normal in $S_4$ – o0BlueBeast0o Apr 26 '14 at 00:36