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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 $3$.

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3 Answers 3

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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?

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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 $$\Bbb Z/6\Bbb Z\qquad\text{and}\qquad\Bbb Z/4\Bbb Z.$$

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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.

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

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