# For which of the following values of n is there a group of order n with no proper normal subgroups?

1) n=21 2) n=9 3) n=60 4) n=98

As per the link here http://groupprops.subwiki.org/wiki/Subgroup_of_index_equal_to_least_prime_divisor_of_group_order_is_normal there is least prime divisor of group order existing for each of the options. Given n=21 we can say that if there is a sub group of order 3 its normal. But how do we prove that a subgroup order 3 exists? With lagrange's theorem we can say that if there is a subgroup of size 3 it is proper subgroup (3 divides 21) but lagrange's theorem cannot be used to prove existence of the subgroup. Can you help me how to find the right answer?

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Do you know Sylow's theorems? –  Daniel Montealegre Mar 21 '12 at 4:46
The Sylow Theorems guarantee the existence of a subgroup of order $p^n$ for any prime $p$ and any positive integer $n$ such that $p^n$ divides the order of $G$. Cauchy's Theorem also guarantees the existence of a subgroup of order $p$ for every prime $p$ that divides $|G|$. See this question for some other "partial converses" to Lagrange's Theorem. –  Arturo Magidin Mar 21 '12 at 4:48
Note that you are misinterpreting the link: the link says that a subgroup whose index is the smallest prime that divides $G$ will be normal. For a group of order $21$, that would be a subgroup of order 7, not $3$. So you would need to guarantee the existence of a subgroup of size $7$, not one of size $3$. But if you don't know about Cauchy's Theorem, then I suspect the truth of that theorem is also beyond what you have on hand... –  Arturo Magidin Mar 21 '12 at 4:49
First, the subgroup of one element is generally considered proper, and always normal, so the question needs restatement. Second, there is a theorem stronger than Lagrange but weaker than Sylow which states that if a prime $p$ divides the order of a group then there is a subgroup of order $p$. Third, are you familiar with any famous examples of nonabelian simple groups? –  Gerry Myerson Mar 21 '12 at 4:50

Well, it is well known that there is a unique simple group of order $60$ upto isomorphism. This group is called the alternating group on $5$ symbols which is denoted by $A_5$.

I am surprised, you pose this as a multiple choice question because the fact I had written above takes a good deal of work.

Also, your contention that the subgroup of order $3$, (if it exists, you're unsure it does, but it actually does!) is normal.

The theorem states that Subgroup of index equal to least prime divisor of group order is normal.

So, the least prime dividing the order of the group is $3$ which must be the index of the subgroup. So, you're looking for the subgroup of order $7$.

To prove the existence of a subgroup of that order, use Cauchy's Theorem.

Similarly, a group in which every element has order divisible by $p$ for a prime $p$ is called a $p$-group. A $p$-group has non-trivial center. Use these fact to conclude that a group of order $9$ has a non-trivial normal subgroup.

Finally for a group of order $98$, use the fact that Sylow $7$-subgroup is order $49$ and hence of index $2$ and hence normal in the whole group.

No, not quite. You seem to think that existence of a subgroup of some order means existence of normal subgroup of that order. But it is not so. You need to prove that the group of order $7$ you got by Cauchy's Theorem is normal. It is by the theorem you have stated. And, the reason you give for $(3)$ is flawed in many ways. I suggest going through more closely. –  user21436 Mar 21 '12 at 7:34