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Find a homomorphism $\phi$ from $U(30)$ to $U(30)$ such that $\ker(\phi) = \{1,11\}$ and $\phi(7) = 7$.

Note: I have seen this Homomorphism from U(30) to U(30) with a given kernel but I have more general questions about the First Isomorphism Theorem that I would like to address. This answer also feels case specific in that it uses the generators, and I am more interested in the general process.

Note 2: We have not covered rings, only groups.

I am having some confusion with the First Isomorphism Theorem, and I want to confirm/deny how I am supposed to interpret this. The statement given in the book is:

Let $\phi$ be a homomorphism from $G$ to $\bar{G}$. Then the mapping from $\frac{G}{\ker(\phi)} \to \phi(G)$ given by $g\ker(\phi) \to \phi(g)$ is an isomorphism. In symbols, $\frac{G}{\ker(\phi)} \approx \phi(G)$

Now, so far in the problems I've done, and the examples we did in class, $\phi(G) = \bar{G}$ has been used. As soon as I got to this problem though, this claim no longer seems to make sense. The kernel is non-trivial, so having $\frac{U(30)}{\{1,11\}} \approx U(30)$ makes no sense because the orders would not match up.

What I am hoping to confirm is how to find $\phi(G)$ because no examples were done in class nor in the book. I would also like to know when it is appropriate to take $\phi(G) = \bar{G}$ so that I avoid this confusion again.

It says the map is given by $g \ker(\phi) \to \phi(g)$. I suppose what I would do is find this map $\forall g \in G$. So, in $U(30) = \{1,7,11,13,17,19,23,29\}$, I would have

$\phi(1) = 1 * \{1,11\} = \{1,11\}$

$\phi(7) = 7 * \{1,11\} = \{7,77\} = \{7,17\}$

$\phi(11) = 11 * \{1,11\} = \{11,121\} = \{11,1\} = \phi(1)$

$\phi(13) = 13 * \{1,11\} = \{13,143\} = \{13,23\}$

$\phi(17) = 17 * \{1,11\} = \{17,187\} = \{17,7\} = \phi(7)$

$\phi(19) = 19 * \{1,11\} = \{19,209\} = \{19,29\}$

$\phi(23) = 23 * \{1,11\} = \{23,253\} = \{23,13\} = \phi(13)$

$\phi(29) = 29 * \{1,11\} = \{29,319\} = \{29,19\} = \phi(19)$

Then, $\phi(U(30)) = \{1,7,13,19\}$

Is this the correct process? If so, is there not an issue with the fact that the result is isomorphic to $\mathbb{Z}_4$ (i.e. isomorphic to a cyclic group)?

From here though, I could claim that $\phi$ is the following:

$\{1,11\} \to 1$

$\{7,17\} \to 7$

$\{13,23\} \to 13$

$\{19,29\} \to 19$

I am having a hard time believing this is indeed a homomorphism though. I feel like this maps to $\phi(U(30))$, not $U(30)$ itself. Can someone convince me that this is an homomorphism?

Thanks again for your help!

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  • $\begingroup$ Which book are you referring to? $\endgroup$
    – Shaun
    Oct 10, 2021 at 17:51
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    $\begingroup$ @Shaun this is from Gallian, Contemporary Abstract Algebra $\endgroup$
    – Nolan P
    Oct 10, 2021 at 18:03
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    $\begingroup$ @GEdgar We have not covered rings, sorry about that. Let me add that to clarify $\endgroup$
    – Nolan P
    Oct 10, 2021 at 18:10
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    $\begingroup$ Can you use (or have you learned) the fact that every finite abelian group is isomorphic to some direct product of finite cyclic groups? (In particular, $U(30)$ has to be isomorphic to $\mathbb{Z}_8$, $\mathbb{Z}_4 \times \mathbb{Z}_2$, or $\mathbb{Z}_2^3$, and it's not hard to figure out which.) $\endgroup$ Oct 10, 2021 at 18:34
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    $\begingroup$ @ConnorHarris yes I can, but I was told I don't even need the abelian assumption. I can just say that $U(n * m) \approx U(n) \oplus U(m)$ where $gcd(n,m) = 1$. In this case, I can write $U(30) \approx U(2) \oplus U(5) \oplus U(3) \approx \mathbb{Z}_1 \oplus \mathbb{Z}_4 \oplus \mathbb{Z}_{2} \approx \mathbb{Z}_4 \oplus \mathbb{Z}_2$. This is actually what I tried first believe it or not, but it didn't get me anywhere. $\endgroup$
    – Nolan P
    Oct 10, 2021 at 18:47

1 Answer 1

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Well we see that $\phi(7)=7\implies\phi(7^{2})=7^{2}$ So $\phi(19)=19$

Similarly $\phi(7^{3})=\phi(13)=13$.

So we know the images of $1,11,7,13,19$.

$\phi(11.7)=\phi(11).\phi(7)=7$

So $\phi(17)=7$. (We are using the fact that $\phi(11)=1$ and $11.7=17)$

again $\phi(11.19)=\phi(11)\phi(19)=19$.

So $\phi(29)=19$. (We are using the fact that $\phi(11)=1$ and $11.19=29)$

Again $\phi(11.13)=\phi(11)\phi(13)$

So $\phi(23)=13$. (We are using the fact that $\phi(11)=1$ and $11.13=23)$

And you have your homomorphism

Namely $\phi$ is the homomorphism which takes

$1\to 1\,,7\to 7\,,11\to 1\,,13\to 13\,,17\to 7\,, 19\to 19\,, 23\to 13\,,29\to 19$

Also since all the information have been generated by the kernel and the known image of the element $7$ this is unique . Now we have to just basically check that this is indeed a homomorphism( it is obvious that it is....but for a complete proof we do have to just do the laborous task and verify it ). You can try and device a computer program for this. I tried to work on something like that once....but I could not finish it.

To answer to your question in the comments:-

you will see that for finite sets of same cardinality a surjection is a injection. So if $\phi(G)=\bar{G}$. You will have that $\phi$ is an isomorphism. In that case the kernel will be trivial. However when you are given that the kernel is non-trivial. Then you have to have that the inamge will only have as many elements as $|\frac{G}{\ker(\phi)}|$. Ofcourse I am only talking about finite groups. So if we were to only use these concepts of First isomorphism Theorem I will argue like this:-

We are given that $\phi(7)=7$ and $\ker(\phi)=\{1,11\}$. So as $\frac{G}{\ker(\phi)}\cong \phi(G)$.

$\phi(G)$ can only have $4$ elements.

But since $\phi(7)=7$,$\phi(7^{2})=7^{2}$ , $\phi(7^{3})=7^{3}$ and $\phi(7^{4})=7^{4}$. We already have $4$ distinct elements.

So the images of the remaining elements must be one of the above $4$. We are given $\phi(11)=\phi(1)=1$. So we are left with $3$ other elements. Now the results have to be such that they agree with the facts $\phi(7)=7$ and $\phi(11)=1$. But as I showed above, the only way to have that is by the way I did . Hence we have our argument.

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  • $\begingroup$ +1 For a clear answer assuming no more theory than necessary. One might want to verify that the map you have determined is indeed a homomorphism; this answer shows that if there is such a homomorphism then it must be this map. But a priori there may be no such homomorphism at all. $\endgroup$ Oct 10, 2021 at 19:29
  • $\begingroup$ Yeah you're right. I verified uniqueness...but for existence part we do have to actually sit and verify that this is a homomorphism. I'll edit it. $\endgroup$ Oct 10, 2021 at 19:34
  • $\begingroup$ @Mr.GandalfSauron I agree that this answers the part of my question that this is indeed a homomorphism. I am satisfied with that. I also found that the comments from earlier verified that I found it correctly. Are you able to answer the remaining part? When is it sufficient to take $\phi(G) = \bar{G}$? $\endgroup$
    – Nolan P
    Oct 10, 2021 at 23:14
  • $\begingroup$ You don't have to have $\phi(G)=\bar{G}$ in order to use first isomorphism theorem. You can say that $\frac{G}{ker(\phi)}\cong \phi(G)$ where $\phi(G)$ is viewed as a subgroup of $\bar{G}$. You will see that this notion will be used several times in gallian. For example in Normalizer-Centralizer theorem. Or Cayley's theorem or even Generalized Cayley. $\endgroup$ Oct 11, 2021 at 6:42
  • $\begingroup$ Also you will see that for finite sets a surjection is a injection. So if $\phi(G)=\bar{G}$. You will have that $\phi$ is an isomorphism. In that case the kernel will be trivial. However when you are given that the kernel is non-trivial. Then you have to have that the inamge will only have as many elements as $|\frac{G}{ker(\phi)}|$. Ofcourse I am only talking about finite groups. Wait I'll edit in these parts. $\endgroup$ Oct 11, 2021 at 6:48

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