Does this group with prime order elements exist?

Question

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.

Answer 1

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 ...

Answer 2

Let $a\in G,\land |a|=2$, and $N=\{e,a\}$ is the unique subgroup with order $2$, hence $N$ is normal in $G$. Let $H$ is another subgroup of $G$, and $|H|=p>2$, where $p$ is prime. If we pick up an arbitrary non-trivial element $h\in H$ and $h\neq e$, since $N$ is normal, we get

$$ah=ha$$

It implies the order of $ah$ must be even (see proof below) and $\ge6$, hence it is a composite number. We get contradictions.

---

Assume the order of $ah$ is odd,

$$(ah)^{2k+1}=e=a^{2k+1}h^{2k+1}=ah^{2k+1}\Longrightarrow h^{2k+1}=a$$

This is impossible, since either $|h^{2k+1}|=p>2$ or $|h^{2k+1}|=|e|=1$, but $|a|=2$

Answer 3

Let $p$ be the smallest prime such that $G$ has an element of order $p$, and let $H$ be the subgroup of order $p$. Notice that $Aut(H)$ has order $p-1$, so $C_G(H)=N_G(H)$, and therefore, since $H$ is normal (as it is the only subgroup of order $p$) it is central. You already know the group is centreless (unless $G=H$), and we obtain a contradiction.