Determining if A is Isomorphic to B

Question

I recently came across this problem:

Consider $\mathbb{Z}_8^*$, which are all the elements of $\mathbb{Z}_8$ that has a multiplicative inverse. So, $\mathbb{Z}_8^*$ = {1, 3, 5, 7}. Now also consider $\mathbb{Z}_{10}^*$ = {1, 3, 7, 9}, and $\mathbb{Z}_{12}^*$ = {1, 5, 7, 11}. Is $\mathbb{Z}_8^*$ isomorphic to $\mathbb{Z}_{10}^*$ or $\mathbb{Z}_{12}^*$? Is $\mathbb{Z}_{10}^*$ isomorphic to $\mathbb{Z}_{12}^*$?

I have discovered that $\mathbb{Z}_8^*$ and $\mathbb{Z}_{12}^*$ are abelian, while $\mathbb{Z}_{10}^*$ is not. Also, I can see that $\mathbb{Z}_8^*$ and $\mathbb{Z}_{10}^*$ are not isomorphic. Is it safe to say that for this case, abelian groups cannot be isomorphic to non-abelian groups? If not, how should I approach this problem?? Thank you.

I drew a cayley table for Z10*, and it wasn't symmetric about the diagonal. — Max, Sep 12 '16 at 03:35
The group operation on $\mathbb{Z}_{10}^*$ ultimately comes from multiplication of integers, which is commutative. So something has gone wrong. To solve the stated problem, try determining how many elements of order $4$ each group has. — carmichael561, Sep 12 '16 at 03:48
If you found the multiplication table for $\Bbb Z_{10}^*$ to be not symmetric about the diagonal, you made a mistake.What you should have found is that the diagonal is not all $1$’s. — Lubin, Sep 12 '16 at 03:56
You're right, I made a mistake finding remainders. Z10* most certainly is abelian. — Max, Sep 12 '16 at 05:58

score 1 · Answer 1 · answered Sep 12 '16 at 04:12

Your three examples are all abelian groups of order $4$. Up to isomorphism, there are only two such groups: $\mathbb{Z}_4$, which is cyclic, and $\mathbb{Z}_2 \times \mathbb{Z}_2$, which isn't. So, which of these two model groups is each of your examples isomorphic to? Every element of $(\mathbb{Z}_8)^{\times}$ has order $2$, so it's isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$. The same is true of $(\mathbb{Z}_{12})^{\times}$. On the other hand, $(\mathbb{Z}_{10})^{\times}$ is cyclic: both $3$ and $7$ are generators (i.e., have order $4$).

Your initial question has an obvious generalization: When are $(\mathbb{Z}_m)^{\times}$ and $(\mathbb{Z}_n)^{\times}$ isomorphic? Clearly one needs the Euler $\phi$-function to agree (i.e., $\varphi(m) = \varphi(n)$) but beyond that... I don't know!

score 0 · Answer 2 · answered Sep 12 '16 at 03:27

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If a group $(G,\cdot)$ is abelian, while a group $(H,\ast)$ is non-abelian, then $G$ and $H$ are not isomorphic. Suppose to the contrary that $\phi: G\to H$ is an isomorphism. Then, $\phi$ is a homomorphism. For $\phi$ to be well-defined, if $c \in G$ and $d\in G$ are such that $c = d$, then $\phi(c) = \phi(d)$. However, $a\cdot b = b\cdot a$, yet $$ \phi(a\cdot b) = \phi(a)\ast\phi(b) \ne \phi(b)\ast\phi(a) = \phi(b\cdot a), $$ in general since $(H,\ast)$ is not abelian. Thus, no such $\phi$ can be well-defined and thus $G$ and $H$ are not isomorphic as groups.

answered Sep 12 '16 at 03:27

Alex Ortiz

24,844

None of his groups are nonabelian. – Lubin Sep 12 '16 at 03:57
@Lubin be that as it may, the OP does ask if abelian groups can be isomorphic to non-abelian groups. – Alex Ortiz Sep 12 '16 at 04:00

Determining if A is Isomorphic to B

2 Answers2