# Does this group with prime order elements exist?

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.

• Let $a$ have order 2 and $b$ have order 3. Suppose $ab$ has order $p$. Then $(ba)^{p+1} = b (ab)^p a = ba$, so $(ba)^p=e$. and $ba$ has order $p$ as well, i.e. both $ab$ and $ba$ are contained in the unique subgroup of order $p$. But $(ba)(ab) = b(a^2)b = b^2$, so $b^2$ is contained in that same subgroup, and hence so is $b$. This means that $ab$ and $ba$ both in fact have order 3, so $ab=ba=b^2$ (since none of those equal $b$ or $e$). But this implies that $a=b$. Commented Jul 7, 2023 at 14:24
• @HaydnGwyn Thats actually an answer. Commented Jul 7, 2023 at 14:42

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 ...

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 $$|a|=2$$

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.