# Let $G$ be finite group and order of $Aut(G)$ is prime number $p$ then prove that order of $G$ is always less than equal to $3$?

As order of $$Aut(G)$$ is prime no. Then this implies $$Aut(G)$$ is cyclic this means $$Aut(G)$$ is abelian this implies inner automorphism group is also cyclic, as cyclic subgroup of cyclic group is cyclic hence as $$Inn(G)$$ is isomorphic to $$G/Z(G)$$. And as $$G/Z(G)$$ is cyclic therefore $$G$$ is abelian. Then how can we say group order cannot exceed $$3$$.

Some hints: since $$G$$ must be abelian, the map $$g \mapsto g^{-1}$$ gives rise to an automorphism of order $$2$$. Hence either this map is the trivial one, that is $$g^2=1$$ for all $$g \in G$$, and $$G$$ must be a direct product of copies of $$C_2$$. Or $$p=2$$. Can you finish?

We can't.

So far, you know that $$G$$ is a finite abelian group, so is the direct sum $$\tag1G=\bigoplus_i C_i$$ of cyclic groups $$C_i=\Bbb Z/\Bbb p_i^{a_i}\Bbb Z$$ of prime power order ($$a_i\ge 1$$, $$p_i$$ prime). Note that $$C_i$$ has $$\phi(p_i^{a_i})=p_i^{a_i-1}(p_i-1)$$ automorphisms and already by combining these, we obtain a subgroup of $$\operatorname{Aut}(G)$$ of order $$\tag2 \prod_ip_i^{a_i-1}(p_i-1).$$ As $$\operatorname{Aut}(G)$$ has prime order $$p$$, $$(2)$$ must be either $$1$$ or $$p$$. In particular, each $$p_i^{a_i-1}(p_i-1)$$ is either $$1$$ or $$p$$. The former happens only for $$a_i=1$$ and $$p_i=2$$, the latter for $$a_i=1$$ and $$p_i=p+1$$ (so $$p=2$$ and $$p_i=3$$) or $$p_i=2$$ and $$a_i=2$$ (so $$p=2$$ again). Hence each $$C_i$$ is either $$\Bbb Z/2\Bbb Z$$ or $$\Bbb Z/3\Bbb Z$$ or $$\Bbb Z/4\Bbb Z$$. So $$G=(\Bbb Z/2\Bbb Z)^a\oplus (\Bbb Z/3\Bbb Z)^b\oplus(\Bbb Z/4\Bbb Z)^c.$$ By the results above, together with permuting equal summands, we already construct a subgroup of $$\operatorname{Aut}(G)$$ or order $$2^b2^ca!b!c!$$. This must be $$1$$ or $$p$$. So the only allowed values for $$(a,b,c)$$ are $$(0,0,0), (0,0,1), (0,1,0), (1,0,0), (1,0,1), (1,1,0), (2,0,0).$$ Among the corresponding groups $$G$$, only the following have order $$>3$$: $$\Bbb Z/4\Bbb Z, \quad \Bbb Z/2\Bbb Z\oplus\Bbb Z/4\Bbb Z, \quad\Bbb Z/6\Bbb Z, \quad(\Bbb Z/2\Bbb Z)^2.$$ Of these, $$\operatorname{Aut}((\Bbb Z/2\Bbb Z)^2)$$ has order $$6$$, $$\operatorname{Aut}(\Bbb Z/6\Bbb Z)$$ has order $$2$$, $$\operatorname{Aut}(\Bbb Z/2\Bbb Z\oplus\Bbb Z/4\Bbb Z)$$ has order $$8$$, $$\operatorname{Aut}(\Bbb Z/4\Bbb Z)$$ has order $$2$$.

So we conclude that the only counter-examples to the desired claim are $$\Bbb Z/6\Bbb Z\qquad\text{and}\qquad\Bbb Z/4\Bbb Z.$$

If the group order exceeds $$3$$, then $$G$$ is cyclic, in which case the automorphism group has order $$\varphi(n)$$, which is never prime, unless $$n=4$$ or $$6$$. For if $$G$$ were not cyclic, it's automorphism group would not be abelian.

So $$G$$ could be $$C_4$$ or $$C_6$$, but those are the only exceptions.

• How are $\phi(4)$ and $\phi(6)$ not prime? Apr 10, 2020 at 10:20
• @HagenvonEitzen I"ll have to correct for that. Thanks.
– user403337
Apr 10, 2020 at 10:21