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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?

codejammer
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  • Do you know Sylow's theorems? – Daniel Montealegre Mar 21 '12 at 04:46
  • 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 04:48
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    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 04:49
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    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 04:50

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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.

Use Google. Good Luck.

(Ping me in case you have some problem.)

  • Using cauchy theorem for possible sub group primes are 1) 1,3,7 2)1,3 3)1,2,3,5 4) 1,2,7 Then from sylow theorem for each we have 1) 7^13=21 so normal subgroup of size 7 exists 2)3^2=9 so normal subgroup of size 3 exists 3)no p exists such that p^nm=|G| where p does not divide m. So 60 has no normal proper subgroup 4)7^2*2=98 so 7 is proper normal subgroup. Is this right? – codejammer Mar 21 '12 at 05:49
  • 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. –  Mar 21 '12 at 07:34
  • can u tell me how to check if the subgroup is normal? – codejammer Mar 21 '12 at 09:10
  • @codejammer There are several ways to do if you know some algebraic details. But, here it is merely number crunching. So, you must have some really deep theorems to tell you that. Sylow's Theorem in one such. –  Mar 21 '12 at 10:29
  • By http://groupprops.subwiki.org/wiki/Subgroup_of_index_two_is_normal we have for 1) 37 index is one. So it is not normal 2)3^2=9 index is 2 and is normal 3)2^215=60 index is 2 and is normal and 4) 7^2*2 again index is 2 and is normal. Is this right? Is index of the group the power we express over p? – codejammer Mar 21 '12 at 11:05
  • No, index of a subgroup is $H$ is the number $|G/H|$... I seriously suggest spending non-trivial amount of time with this. It is not something you learn over $3$ comments. Please... –  Mar 21 '12 at 12:05
  • Yes I got it now, we need to use cacuhy's theorem and list all prime factors. Expand to all possible factors by using Sylow's (various P^n factoring out that which do not divide G by Lagranges) and check for each of these numbers whether they are normal by computing their index (diving G by size of H) and they are normal if their index is 2 or index is equal to the least prime divisor. Phew been a long day. – codejammer Mar 21 '12 at 12:56
  • You're probably right now, but I am not quite satisfied with your description. –  Mar 21 '12 at 12:57
  • And, by no means, these are the only ways to prove normality. There are quite a lot of ways to prove... –  Mar 21 '12 at 12:58
  • Thanks, Can you suggest some good reading on this topic? Some text that has lot of examples perhaps... – codejammer Mar 21 '12 at 13:01
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    Have you looked at Dummit and Foote, Abstract Algebra? –  Mar 21 '12 at 13:06
  • @codejammer: You are misinterpreting what you are reading again. The result you quote says "If the index of $H$ is $2$, then $H$ is normal." It does not say "if the index of $H$ is not $2$ then $H$ is not normal." Any subgroup of index $1$ equals the whole group, so it is normal. – Arturo Magidin Mar 21 '12 at 14:37
  • @ArturoMagidin yes but since I am looking for proper subgroups as per my question I should be looking at groups with index greater than 1 right? – codejammer Mar 22 '12 at 07:15
  • @codejammer Yes, you're right. But, he is trying to point out that, you seem to misunderstand. My very humble suggestion, please take a look at Dummit and Foote. Their Examples will help a lot in clearing your ideas. You may also want to look at some of the answers that give arguments about a subgroup of particular order is not simple. For instance, Arturo has written a lot of such of them. Another user named m.k. has also written quite a few. I have written a few myself. –  Mar 22 '12 at 07:38
  • @KannappanSampath yes I have started reading it, I am finding it easier to understand than Topics in Algebra - I.N. Herstein which I was reading earlier. – codejammer Mar 22 '12 at 10:08
  • @codejammer Glad to hear that. You could drop into chat and leave me a message if you find something hard to grasp at first go. When I get notified, I'll try to clarify. –  Mar 22 '12 at 10:13
  • @JyrkiLahtonen Thank You. –  Apr 16 '12 at 05:08