What group is $(\mathbb{Z}/24\mathbb{Z})^{*}$ isomorphic to I want to determine which group $(\mathbb{Z}/24\mathbb{Z})^{*}$ is isomorphic to.
$\mathbb{Z}/24\mathbb{Z}$ contains the 24 residue classes $z + 24\mathbb{Z}$ of the division mod 24. For brevity, I will identify them with $z$, so $\mathbb{Z}/24\mathbb{Z} = \{ 0, 1, ..., 23\}$. For $(\mathbb{Z}/24\mathbb{Z})^{*}$, I have the definition $\{z \in \mathbb{Z}/24\mathbb{Z}\,\,|\,\, \exists z^{-1}\in\mathbb{Z}/24\mathbb{Z}: zz^{-1}=1\}$ with the usual multiplication. So $((\mathbb{Z}/24\mathbb{Z})^{*}, \times)$ is a group.
What I get is $(\mathbb{Z}/24\mathbb{Z})^{*} = \{1,5,7,11,13,17,19,23 \}$. My reasoning is this: $0$ is out because we can't satisfy $0z=1$. Therefore, all divisors of 24 are out. $9,15,16,20,21$ violate the uniqueness of the neutral element because they don't change by squaring. $10,14,18,22$ yield divisors of $24$ by squaring. This is kind of clumsy. Is there a faster way to notice that only the primes are contained in $(\mathbb{Z}/24\mathbb{Z})^{*}$?
In order to determine which group $(\mathbb{Z}/24\mathbb{Z})^{*}$ is isomorphic to, I calculated the order of each element which is two. Again, is there an easy way to see that primes are always of order two in such groups?
Another group, where all elements are of order two is $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$. I conjecture that two groups are isomorphic if (and only if) their elements are of the same order but I don't see how to prove this. In my special case, I could calculate the Cayley table of both groups and define the homomorphism by explicitly stating which element gets assigned to which but maybe there's an easier and more general way to prove this?
I already got a hint that I should use the fact that there are 16 subgroups of $(\mathbb{Z}/24\mathbb{Z})^{*}$ but I don't see where this gets me unless I can prove that $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$ is the only group of order 8 with 16 subgroups also.
 A: Every finite abelian group is a product of cyclic groups. If every element has order two, then it has to be a product of copies of $\mathbf{Z} / 2$.
I believe your conjecture is true for finite abelian groups, but false for general finite groups (and false for infinite abelian groups). But it's been a long time since I've looked at that problem.
Incidentally, the usual approach is to factor into prime powers: by the Chinese Remainder Theorem,
$$ (\mathbf{Z}/24)^\times \cong (\mathbf{Z}/8)^\times \times (\mathbf{Z}/3)^\times$$
and if you've seen the unit groups for prime powers characterized, you can just plug that in:


*

*$(\mathbf{Z} / 2)^\times \cong 0$

*$(\mathbf{Z} / 2^n)^\times \cong \mathbf{Z}/2 \times \mathbf{Z}/2^{n-2}$ if $n > 0$

*$(\mathbf{Z} / p^n)^\times \cong \mathbf{Z}/p^{n-1} \times \mathbf{Z}/(p-1)$ if $p$ is an odd prime

A: First, the fact that $\left(\mathbb{Z}/24\mathbb{Z}\right)^\times$ contains exactly the primes is an accident of the number $24$. For example, $(\mathbb{Z}/32\mathbb{Z})^\times$ contains lots of non-primes ($9$, $15$ for example). The elements of $(\mathbb{Z}/n\mathbb{Z})^\times$ are exactly the integers less than $n$ and prime to it, of which there are $\phi(n)$ where $\phi$ is the Euler phi function.
You can indeed define an explicit homomorphism from your group to $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$. For example, send $5\mapsto (1,0,0)$ and $7\mapsto (0,1,0)$. That determines the image of $5\times 7 = 11$; then you can for example send $13\mapsto (0,0,1)$. You still have the problem of proving that this is an isomorphism.
If you know the fundamental structure theorem for abelian groups, you can see that $\left(\mathbb{Z}/24\mathbb{Z}\right)^\times$, being an abelian group of order $8$, must be isomorphic to $\mathbb{Z}/8\mathbb{Z}$, $\mathbb{Z}/4\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$, or $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$. The fact that all elements are of order $2$ implies that the last one must be the case.
Alternatively, as Hurkyl points out, there are explicit formulas for this group for arbitrary $n$.
A: Very nicely posed question, and some good work on your part. For the first part, identifying the elements of $(\mathbb{Z}/24\mathbb{Z})^{\star}$, note that an element of $\mathbb{Z}/24\mathbb{Z}$ is a unit if and only if it is relatively prime to $24$. You can see this by noting that if $\gcd(\alpha, 24) = 1$, then there exist $\beta, \gamma \in \mathbb{Z}$ such that $\alpha\beta + 24\gamma = 1$. Now reduce mod $24$. Similarly, if $\alpha \in (\mathbb{Z}/24\mathbb{Z})^{\star}$, then there exists $\beta \in \mathbb{Z}/24\mathbb{Z}$ such that $\alpha\beta \equiv 1 \pmod {24}$, so there exists $k \in \mathbb{Z}$ such that $\alpha\beta + 24k = 1$, i.e. $\gcd(\alpha, 24)=1$. Note that this fact can be generalized to $\mathbb{Z}/n\mathbb{Z}$ for any $n \in \mathbb{N}$: the units of $\mathbb{Z}/n\mathbb{Z}$ are precisely the elements of $\mathbb{Z}/n\mathbb{Z}$ relatively prime to $n$.
As far as your second question goes, the image of an element of order $n$ under an isomorphism is again an element of order $n$. That is, if $\phi:G\rightarrow H$ is an isomorphism, then if $|g| = n$ for some $g \in G$, then the order of $\phi(g)$ is also $n$. This is because $1_{H} = \phi(1_{G}) = \phi(g^{n}) = \phi(g)^{n}$, so the order of $\phi(g)$ divides $n$. Now suppose $\phi(g)^{k} = 1_{H}$ for some $k$. Then this implies $\phi(g)^{k} = \phi(g^{k}) = 1$, i.e. $g^{k} \in \ker(\phi)$. But $\ker(\phi) = \{1_{G}\}$, since $\phi$ is an isomorphism, so $g^{k} = 1$, i.e. $n | k$. Hence, isomorphisms preserve orders of group elements. 
Hence, if every element of $(\mathbb{Z}/24\mathbb{Z})^{\star}$ has order $2$, then its isomorphism type must be a group of order $8$ where every element has order $2$. The only such candidate is indeed $\mathbb{Z}_{2}\times \mathbb{Z}_{2} \times \mathbb{Z}_{2}$. In general, Hurkyl's observation is the best way to tackle this particular type of problem, but I don't know if you have seen either the Chinese Remainder Theorem, or the Fundamental Theorem of Finitely Generated Abelian Groups, so I thought I'd offer a slightly different perspective. 
