Group of order 9 is simple 
Is't true that a group of order $9$ is simple? 

How can it be proved or disproved. 
 A: When we say "A group of order $9$ is simple" we say that every group of order $9$ is simple.
To disprove this we only need to find one counterexample, that is one group which has $9$ elements and it is not simple. 
Remember that an abelian group is simple if and only if it is trivial, or its order is prime. Can you find an abelian group of order $9$? Is $9$ a prime number?
A: No. You can use the first Sylow theorem which says that if $G$ is a group and the order of $G$ which we denote $|G|$ is of the form $p^nm$. Then $G$ contains a subgroup of order $p^i$ normal in a subgroup of order $p^{i+1}$ for all $i < n$. In our case $|G| = 9 = 3^2$ so G has a subgroup of order $3 = 3^1$ which is normal in a subgroup of order $3^{1+1} = 3^2 = 9$ hence this subgroup of order $3$ is normal in all of $G$. So $G$ is not simple.
A: Hints - let $\,G\,$ be a group of order $\,9\,$ , then
1) Show $\,G\,$ is abelian, for example by using (showing) that $\, |Z(G)|>1\,$
or
2) As with any other finite $\;p$-group , show $\,G\,$ has a normal subgroup of prime order
or
3) Use Cauchy's Theorem to show $\,G\,$ has an element of order $\,3\,$ that generates a normal subgroup (why normal? Because its index in $\,G\,$ is the minimal prime that divides $\,|G|\,$)
The above are just some of the basic approaches (pretty close to each other, btw) you can use depending on how advanced in your studies you are. Your turn now. 
A: Not, it is not. It is a general fact that if $p$ is a prime then every group of order $p^n$ with $n\geq 1$ has non trivial center. Moreover, you can prove that every group of order $p^2$ is abelian. This follows from the following exercise: if $G$.is a group such that $G/Z(G)$ is cyclic, then $G$ is abelian.
A: If $|G|=p^2$ then $G$ is abelian. To see this, you should prove that if $G/Z(G)$ is cyclic then it is trivial. This is a classis question which can be found in standard texts on group theory, or on math.stackexchange!
Thus, can $G$ be simple?
