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

How can it be proved or disproved.

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2  
Several of us are unsure whether you want to know whether there exists a simple group of order 9, or whether all groups of order 9 are simple. Could you clarify please? (But the answer is no to both questions.) –  Derek Holt Apr 10 '13 at 14:24

5 Answers 5

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.

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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?

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Surely the OP is just getting slightly confused and is meaning to ask "Is there a simple group of order $9$?"? –  user1729 Apr 10 '13 at 10:58
    
    
@user1729: On a less Leslie Nielsen reply, I find this question to actually make some sense as posed. Especially since I don't know what does the OP know or don't know about groups. –  Asaf Karagila Apr 10 '13 at 11:10
    
I know what you mean, but the "$G/Z(G)$ cyclic $\Rightarrow G$ abelian" is a pretty early-on question (pre-Sylow theorems), and, well, why $9$? Why not just "Prove that there is a non-simple group of order $n>1$ if and only if $n$ is composite?"...? –  user1729 Apr 10 '13 at 11:14
    
Personally, I would never write "a group of order 9 is simple" or even "a group of order 3 is simple", because the meaning is ambiguous. The fact that the first statement is false and the second one true under any reasonable interpretation is beside the point. –  Derek Holt Apr 10 '13 at 13:54

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.

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The (or "a") theorem is even stronger: any finite $\,p$-group has a normal subgroup of any order dividing its order. –  DonAntonio Apr 11 '13 at 10:12

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?

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Please have a look at this question. I asked it as you suggested. You made me do that. Thanks –  Babak S. Jun 12 '13 at 19:26

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.

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