Does there exist a group such that each non trivial element has prime order and for each prime $p$ exactly $p-1$ elements are there with order $p$? Such group say $G,$ if exist is obviously infinite. By definition it contains a unique subgroup of each prime order. The group has to be non abelian, in fact the center must be trivial. This is because, if $ab = ba$ and if $a$ has order $p$ and $b$ has order $q$ where $p \ne q$ then $ab$ has order $pq$ which is not possible.
Does this group with prime order elements exist
abstract-algebraexamples-counterexamplesgroup-theoryinfinite-groups
Related Solutions
$G$ does not have to be abelian. There exist examples of infinite (necessarily highly nonabelian) groups with exactly two conjugacy classes; that is, even the action of inner automorphisms on non-identity elements is transitive.
As observed in the comments, since $[G, G]$ is a characteristic subgroup, if $G$ is nonabelian then it is necessarily perfect.
As observed in the comments, nonabelian free groups are not perfect.
I don't know the answer to the last question. It seems potentially difficult due to the existence of Tarski monsters.
Groupprops claims that for abelian groups this property is equivalent to being the underlying additive group of a field, although I don't immediately see a proof. As observed in the comments, as groups these are just vector spaces over $\mathbb{F}_p$ or $\mathbb{Q}$.
Edit: Here's a proof that if $G$ is abelian then it's a vector space over $\mathbb{F}_p$ or $\mathbb{Q}$. As you observed, if $G$ has an element of finite order then every element has order $p$ for some prime $p$, hence $G$ is an $\mathbb{F}_p$-vector space. Otherwise, every element has infinite order. If $g \in G$ and $n \in \mathbb{N}$, then by assumption there is an automorphism $\varphi : G \to G$ such that $\varphi(ng) = n \varphi(g) = g$. Hence $G$ is divisible, and an abelian torsion-free divisible group is a $\mathbb{Q}$-vector space.
Let $\Omega_i(G)$ denote the subgroup generated by all elements of order $p^i$. Recall that an elementary abelian group is simply a direct product of (possibly infinitely many) copies of the cyclic group $C_p$*. Its rank is the number of copies of $C_p$. Notice that raising to the power $p$ is a map from $\Omega_i(G)$ to $\Omega_{i-1}(G)$ for each $i$. In particular, it induces a homomorphism from the quotient groups $\Omega_{i+1}/\Omega_i(G)\to \Omega_i(G)/\Omega_{i-1}(G)$, and this homomorphism must be injective. Since all elements of this quotient have order $p$, it is an elementary abelian $p$-group, so we are interested in its rank. In particular, the rank of $\Omega_i/\Omega_{i-1}(G)$ is at most the rank of $\Omega_1(G)/\Omega_0(G)=\Omega_1(G)$. For $i\in \mathbb N$, write $r(G,i)$ for the rank of $\Omega_i(G)/\Omega_{i-1}(G)$ (which could be $\infty$), a weakly descending sequence for any fixed $G$, i.e., $r(G,i)\geq r(G,j)$ for all $i\leq j$. This applies for all subgroups of $G$ as well, and of course $r(G,i)\geq r(H,i)$ for any subgroup $H$ of $G$.
Thus if there are infinitely many elements of order $p^i$ for some $i$ then there are infinitely many of order $p$. These look like an infinite-dimensional $\mathbb{F}_p$-vector space. Take the set of all infinite subgroups of $\Omega_1(G)$. Certainly there are infinitely many of them, and they cannot have a minimal element, just by removing one basis element at a time. Thus $G$ cannot have min.
Conversely, let $$H_1>H_2>H_3>\cdots$$ be an infinite descending chain of subgroups of $G$, but that $G$ has only finitely many elements of order $p$ (and hence of order $p^i$ for all $i$). For each $i$, let $f(i)$ denote the smallest order of an element of $H_i\setminus H_{i+1}$, which exists since all elements of $G$ have finite order. Notice that $f:\mathbb N\to\mathbb N$ cannot take the same value $m$ infinitely often, as then there would be infinitely many elements of order $p^m$ in $G$.
Note that $r(H_i,j)\geq r(H_{i+i},j)$ for all $i$ and $j$. If $f(1)=m_1$, then we see that $r(H_1,m_1)>r(H_2,m_1)$, which means that the latter is strictly less than $r$, and indeed therefore $r(H_i,m)<r$ for all $i\geq 2$ and $m\geq m_1$. We will now repeat the process, but we have to be a little careful.
Since the sequence $f(i)$ diverges, we can produce a weakly increasing subsequence, with indices $i_1,i_2,\dots$, and values $m_j=f(i_j)$. We have $$ r\geq r(H_{i_1},m_1)>r(H_{i_1+1},m_1)\geq r(H_{i_2},m_2)>r(H_{i_2+1},m_2)\geq r(H_{i_3},m_3)>\cdots.$$ Such a sequence cannot have more than $r$ strict inequalities, and so we obtain a contradiction.
*Here direct product means direct sum in the categorical sense, sometimes a restricted direct product. How about: this is an abelian group all of whose elements have order $p$.
Best Answer
I do not think this can be realized: if $M$ is the unique subgroup order $p$ and $N$ the unique one of order $q$, with $p,q$ different primes, then these are normal subgroups. Obviously $M \cap N=1$, whence $MN \cong M \times N \cong C_{pq}$ and this subgroup of $G$ contains non-trivial elements of non-prime order ...