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So I'm having trouble with this problem. I know that the definition of a simple group means that the group has no nontrivial subgroups. I know that this can be proven somehow with the help of the converse of Lagrange's Theorem for Abelian groups: If G is abelian of order n, and d is a divisor of n, then G has a subgroup of order d.

My attempt: (=>)Assume that G is a finite abelian group and is simple, then G has no nontrivial normal subgroups. (Now I don't know how to show that this implies that G has order p, where p is prime.

(<=)Assume that G is a finite abelian group with order p, where p is a prime. (Since the order of p is prime then what does this mean?)

Edit: Can someone check my new attempt at the proof?

(=>) Suppose G is a simple finite abelian group. Suppose for the sake of contradiction that G does not have prime order, then |G|=p*k where p is a prime number and k is an integer such that k>1. Then G has an element of order p. Let the element of order p be called x. Then , the subgroup generated by x, is of order p and is not all of G. Since G is abelian, this subgroup is normal, which leads us to a contradiction. Therefore, G must have prime order.

(<=) Suppose that G is a finite abelian group and it’s order is p, a prime. Since G has prime order, then the only two subgroups of G are the trivial subgroup and the group G. Then, by definition the group G is simple since there are no nontrivial proper subgroups, and thus no nontrivial normal subgroups.

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marked as duplicate by user26857 abstract-algebra Oct 31 '15 at 22:59

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1  
You might find your answer in the following link. math.stackexchange.com/questions/186035/… – user112535 Dec 7 '13 at 2:26
    
A simple group has no nontrivial normal subgroups. – Josh B. Dec 7 '13 at 2:27
1  
You can omit the word "finite" and this is still true. – Derek Holt Dec 7 '13 at 10:49
    
Simple groups can have proper nontrivial subgroups (they have to be nonabelian though). I think Cauchy's is too high-tech for this: – anon Dec 8 '13 at 1:42
    
Say $G$ is nontrivial abelian and pick a nontrivial $g\in G$ then consider $\langle g\rangle$; it's automatically nontrivial, and if it's proper in $G$ then you're done (subgroups of abelian groups are automatically normal, so here $G$ would be nonsimple), otherwise $\langle g\rangle=G$ and we know we can consider cyclic groups. In $C_n=\langle g\rangle$, if $d\mid n$ is a nontrivial factor then $\langle g^d\rangle$ is proper and nontrivial, hence $G$ is nonsimple when $n$ is composite. As for $C_p$, it has no proper nontrivial subgroups at all, which is easy to check. – anon Dec 8 '13 at 1:47

Edit: Can someone check my new attempt at the proof?

(=>) Suppose G is a simple finite abelian group. Suppose for the sake of contradiction that G does not have prime order, then |G|=p*k where p is a prime number and k is an integer such that k>1. Then G has an element of order p. Let the element of order p be called x. Then , the subgroup generated by x, is of order p and is not all of G. Since G is abelian, this subgroup is normal, which leads us to a contradiction. Therefore, G must have prime order.

(<=) Suppose that G is a finite abelian group and it’s order is p, a prime. Since G has prime order, then the only two subgroups of G are the trivial subgroup and the group G. Then, by definition the group G is simple since there are no nontrivial proper subgroups, and thus no nontrivial normal subgroups.

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Yes this is correct. I feel Cauchy's is sort of high-tech for the level the problem is at though. – anon Dec 8 '13 at 1:50

Hint :

For $\Rightarrow $ :

Suppose group has order $|G|=pqk$ where $p,q$ are distinct primes.

Now, what doe Cauchy theorem tells for finite abelian groups?

For $\Leftarrow $ :

For a group to be simple, It should have some non trivial subgroup.

Order of a subgroup divides order of group.

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If $G$ is a simple finite abelian group, then it has no non-trivial subgroups, since all subgroups are normal because $G$ is abelian. This means that $G$ is cyclic, generated by any non-trivial element $g$. Let $n$ be the order of $G$. If $d$ divides $n$, then $\langle g^d \rangle$ is a subgroup of $G$. Thus, $d$ can only be $1$ or $n$, and so $n$ is prime.

If $G$ is a finite group whose order is prime, then $G$ is cyclic, generated by any non-trivial element. This implies that $G$ has no non-trivial subgroups.

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Your proof seems good. Here's a different approach without using cauchy's theorem.

$(\Longrightarrow)$ If $G$ is a finite simple abelian group, pick any $g \in G$ non trivial. Then $1 \not =<g> \trianglelefteq G$ since $G$ is abelian. Since $G$ is simple, $G = <g>$. That is, $G$ is cyclic. Since $G$ is finite and cyclic, for every divisor $d$ of $|G|$ ,$G$ has a unique cyclic subgroup of order $d$. $G$ is simple so that this cyclic subgroup is either $G$ or 1. Thus the divisor $d$ is either 1 or $|G|$. This show $|G|$ must be prime.

For fun, you can modify the problem a little bit more.

Let $G$ be an simple abelian group, not necessarily finite. Prove that $G$ must be finite and of prime order.

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You can use \langle and \rangle to produce $\langle$ and $\rangle$ instead of $<$ and $>$. – Michael Albanese Nov 19 '15 at 20:48

I have noticed a few issues with the definition of 'simple' in a number of answers and comments. A finite group $G$ is simple it it has no non-trivial $normal$ subgroups. In the case where $G$ is abelian, every subgroup is normal by simply using the definition of normal. Just something to take note of and be carful of.

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(<=) Let $G$ be a prime order abelian group. If G had a subgroup of any order other than $|G|$ or $1$ (say q where $1<q<|G|$) we would conclude that $q$ divides $|G|$ by Lagrange's Theorem, a contradiction since $|G|$ is prime. The subgroups of orders $|G|$ and $1$ are trivial. All of the subgroups of $G$ are trivial, and so all of the normal subgroups are trivial. Therefore $G$ is simple.

(=>) Let $G$ be a finite, abelian, simple group. Assume for a contradiction that $|G|$ is not prime. Since $|G|$ is not prime, we can find $1<d<|G|$ such that $d$ divides $|G|$. By the converse to Lagrange's Theorem described in the question, there is a subgroup of $G$ (call it $H$) such that $|H|=d$. It is clear that $H$ is not $G$ neither $\left\{e\right\}$ (the group consisting of the identity of $G$), since its order differs from the order of these groups.

We have found a non-trivial subgroup. To complete the contradiction, we need to show that it is normal. Finding a normal, nontrivial subgroup would contradict our assumption of the simplicity of G.

$H$ is normal since $G$ is abelian. By definition, all elements of $G$ commute under its binary operation. Fixing $g$ in $G$ we find that: $$\left\{hg : h\in H\right\}=\left\{gh : h\in H\right\}$$ This holds because $g$ and $h$ commute in G for any h in H. We found that: $$gH=Hg; \forall g\in G$$ Hence, by definition, $H$ is normal. This yields a contradiction as outlined above.

We conclude that $|G|$ is prime. //

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protected by user26857 Nov 1 '15 at 18:04

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