I not getting how to start with this.I have completed Lagrange theorem.
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what does lagrange theorem say? – Sep 20 '15 at 04:52
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1Step 1: Prove that a group of order $p^n$ has a nontrivial center. Step 2: Prove that a group with cyclic center is abelian. – Alex G. Sep 20 '15 at 04:52
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@AlexG. : Do you really mean a group with cyclic center is abelian?? – Sep 20 '15 at 04:56
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2Oops, no. Rather, if $G/Z$ is cyclic, then $G$ is abelian – Alex G. Sep 20 '15 at 04:57
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What is "this"? The post doesn't seem to state a problem. – Carl Mummert Sep 23 '15 at 22:04
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possible duplicate of Elementary approach to proving that a group of order 9 is Abelian – Sep 23 '15 at 22:55
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A group of order $p^2$ with p a prime is always commutative :
Because the cardinal is a power of $p$, the centre of $G$ is non trivial : indeed, $|Z(G)|\equiv 0 [p]$, so cannot be $1$.
Then, if $G$ was not abelian, you would have $Z(G)\subset G$ not to be an equality. Like $|Z(G)|$ divides $|G|$, it is $1$, $p$, or $p^2$. It is not $1$ because the centre is non trivial, and it is not $p^2$, because $G$ is not abelian.
So $Z(G)$ has p elements. Then you take $g\in G\setminus Z(G)$. The subgroup generated par $G$ and $g$, is of cardinal greater than $p+1$, and dividing $p^2$, so it is $G$.
Then, every element $y$ of $G$ is $y=z.g^k$, where $z\in Z(G)$, and $k\in \mathbb Z$. But so $yg=zg^{k+1}=zgg^k=gzg^k=gy$ for all $y$. Then $g\in Z(G)$, and that is absurd.
So $G$ has to be abelian.
D.L.
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We know that any group of order $p^2$ where $p$ is a prime is abelian. In your case $9=3^2$, hence the group is abelian.
For more details see the link Any group of order prime square is abelian
Kushal Bhuyan
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