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

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?

• Which book are you referring to? Oct 10, 2021 at 17:51
• @Shaun this is from Gallian, Contemporary Abstract Algebra Oct 10, 2021 at 18:03
• @GEdgar We have not covered rings, sorry about that. Let me add that to clarify Oct 10, 2021 at 18:10
• 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.) Oct 10, 2021 at 18:34
• @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. Oct 10, 2021 at 18:47

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 $$11.13=23)$$

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.

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.

• +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. Oct 10, 2021 at 19:29
• 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. Oct 10, 2021 at 19:34
• @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 $\phi(G) = \bar{G}$? Oct 10, 2021 at 23:14
• 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. Oct 11, 2021 at 6:42
• 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. Oct 11, 2021 at 6:48