Why do we care about elements' order if looking for the isomophism types? I have to exercise several things during the abstract algebra course. Among them is

*

*a theorem about finitely generated abelian groups, which says that for a finitely generated abelian group $G$ there are numbers $s,n \in \mathbb{N}_0$, $d_1, \ldots, d_s \in \mathbb{N}_{>1}$ with $d_i \mid d_{i+1}$ for $i \in \{1, \ldots, s-1\}$ and $G \cong Z_{d_{1}} \times \cdots \times Z_{d_s}\times Z_\infty \times \cdots \times Z_\infty$;

*the fact that $(\mathbb{Z}/ (n))^\times$ consists of all classes $\bar{a}$ so that $gcd(a,n)=1$.

Here $Z_d$ stands for a cyclic group of the order $d$ and $R^\times$ is a unit group of the ring $R$.
Now I have en exercise where I have to determine the isomorphism types for different groups. Reading the solution I do not particularly understand why the order of elements defines the isomorphism types. In particular,

*

*$((\mathbb{Z}/(8))^\times = \{\bar{1}, \bar{3}, \bar{5}, \bar{7}\}$, so the types $Z_2 \times Z_2$ and $Z_4$ are possible. The solutions says, as every element has an order $\le 2$ we have $(\mathbb{Z}/(8))^\times \cong Z_2 \times Z_2$. I am asking myself where is the bridge, why does it matter, why the order determines the isomorphism type, and why $Z_4$ is not possible.


*$(\mathbb{Z}/(10))^\times=\{\bar{1}, \bar{3}, \bar{7}, \bar{9}\}$. The solution says, as $ord(3) > 2$ we have $(\mathbb{Z}/(10))^\times \cong Z_4$. I have the same questions as above.


*$(\mathbb{Z}/(15))^\times = \{\bar{1}, \bar{2}, \bar{4}, \bar{7}, \bar{8}, \bar{11}, \bar{13}, \bar{14}\}$. The solution says that we have three possible types: $Z_2 \times Z_2 \times Z_2, or Z_2 \times Z_4, or Z_8$. I agree on that because that corresponds to the theorem. For the first case all elements must be self-inverse, which does not fit. I agree, $(2)^{-1}=8$, but how does it link to the isomorphism type? In the group $Z_8$ there is only one element of the order $2$, which is $a^4$. This does not fit as well. I agree, we have $(4)^{-1}=4, (11)^{-1}=11$, so there are at least two elements of the order $2$, but how does it link to the isomorphism type? Finally, we have only $Z_2, \times Z_4$ left, which is the isomorphism type.
So, could you please explain me what I am missing and explain me that using these three simple examples? I do not understand the logic and the art behind using the order of an element to determine the isomorphism type of a group. I am sad because I miss some basic stuff which seems to be self-evident for the author of this exercise, but I want to learn that.
 A: Here's a useful little lemma:

If $f : A \to B$ is an isomorphism between two groups, and if $a \in A$ and $b=f(a) \in B$, and if $a$ and $b$ have finite order, then the order of $a$ equals the order of $b$.

Here's the proof. Let $m$ be the order of $a$ and let $n$ equal the order of $b$. It follows that
$$\text{Id}_B = f(\text{Id}_A) = f(a^m) = f(\underbrace{a a \ldots a}_{\text{$m$ times}}) = \underbrace{f(a) f(a) \ldots f(a)}_{\text{$m$ times}} = \underbrace{b b \ldots b}_{\text{$m$ times}} = b^m
$$
and therefore $n \le m$ (of course there's really an induction argument hidden here, in which one proves that $f(a^i)=b^i$ for all $i \ge 1$). Similarly, using the inverse isomorphism $f^{-1}$, one proves that $m \le n$. Therefore $m=n$.
As a corollary, if $A,B$ are isomorphic groups, and if $A$ has no elements of order $4$, then $B$ also has no elements of order $4$. This should settle your $(\mathbb Z / (8))^\times$ example.
I'm sure you can now figure out how to apply similar logic to the other examples (and, please, only one question per post).
And, by the way, the statement of that lemma is unnecessarily weak. If you look at it's proof, you'll see that something stronger is proved, namely you do not need to assume that $a$ and $b$ have finite order. The proof shows that if $a$ has finite order then so does $b$, and if $b$ has finite order then so does $a$, and therefore $a$ has finite oder if and only if $b$ has finite order.
A: *

*If every element of a group $G$ has an order $\le2$, $G$ can't be (isomorphic to) $(\mathbb Z_4,+)$ because in $\mathbb Z_4$ there are elements of order $4$.

*If some element of a group $G$ has an order $>2$, $G$ can't be (isomorphic to) $(\mathbb Z_2,+)^2$ because in $\mathbb Z_2^2$ every element has an order $\le 2$.

*Being not self-inverse is equivalent to having an order $>2$.If a group $G$ has at least two elements of order $2$, $G$ can't be (isomorphic to) $(\mathbb Z_8,+)$ because the latter has only one element of order 2.Btw, the elements of order 2 in $(\mathbb{Z}/(15))^\times$ are not $\bar7^{-1}=\bar{13}$ but $-\bar1=\overline{14}$, $\bar4$ and $-\bar4=\overline{11}$.

