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Let $G$ be a finite group, such that $\mid Aut(G)\mid=p^n$. Then prove $G$ is p-group or $G\cong P\times C_{2}$, where $P$ is a p-group.

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The statement is false. Take $G=C_{30}$. Then $|Aut(C_{30})|=\varphi(30)=8=2^3$. I will change your statement by this one: If $G$ is a finite group such that $|Aut(G)|=p^n$ for some prime $p$ then $G$ is one of the following groups: $P$, $P\times C_3$, $P\times C_5$, $P\times C_{15}$, where $P$ is a 2-group which has not a direct factor of the form $C_2^2$ or $C_{2^k}$ with $k>1$; or if $p\geq 3$ then $G$ is of the form: $Q$ or $Q\times C_2$ where $Q$ is non-abelian $p$ group.

Since $G/Z(G)$ is embedded in $Aut(G)$ then $G/Z(G)$ is a finite $p$-group and therefore $G$ is nilpotent. Then $G=S_{p_1}\times\dots\times S_{p_k}$ where $S_{p_i}$ is the $p_i$-Sylow subgroup of $G$, where $p_1,\ldots,p_k$ are the prime numbers dividing the order of $G$. Then $Aut(G)=Aut(S_{p_1})\times\dots\times Aut(S_{p_k})$.

If $P$ is a non-abelian finite $p$-group then $Z(P)\neq P$ and therefore $Aut(P)$ contains an element of order $p$ because $P/Z(P)$ is embedded in $Aut(P)$. Then at most one of the $S_{p_i}$ is non-abelian.

If $P$ is a finite abelian $p$-group then $P=P_1\times P_2$ with $P_1$ cyclic of maximal order, say $p^s$, and therefore $Aut(P_1)\subseteq Aut(P)$ with $|Aut(P_1)|=\varphi(p^s)=p^{s-1}(p-1)$ which is not a prime number if $p>5$. This also implies that $2$ divides $|Aut(P)|$ if $p\geq 3$ and that $p$ divides $Aut(P)$ for every abelian finite $p$-group $P$ unless $P$ is elemental, that is, if $G$ is of the form $C_p^k$. But $|Aut(C_p^k)|=(p^k-1)(p^k-p)\dots (p^k-p^{k-1})$ which is not prime if $k>1$.

We conclude that if $|Aut(G)|=2^n$ then $G$ is the product of a 2-group $P$ which has not a direct factor of the form $C_2^2$ or $C_{2^k}$ with $k\geq 2$, and a group $Q$ which is one of the following list: $C_3,C_5,C_{15}$. Example $C_2\times C_3\times C_5$. If $|Aut(G)|=p^n$ with $p\neq 2$ then the remarks above say that $G$ is the product of a non-abelian $p$-group $P$ with $1$ or $C_2$.

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@ Diego: Thank you for answer. I forgot condition "odd prime $p$ in question. The question is true for odd prime $p$. –  maryam Jan 18 '13 at 4:00
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