Abstract Algebra question?

Question

Prove : A group of order 2002 cannot be simple.

A group is simple if it has no normal subgroups

Answer 1

The previous poster has the right method, but use 11 as your prime. You then need a divisor of 2*7*13 which is 1 modulo 11. This shows that there is only one 11-Sylow subgroup, so it is normal.

Answer 2

Look at Sylow's theorems in your text. There you will find the following facts:

For any prime p dividing the order n of the group, the number k_p of Sylow p-subgroups is congruent to 1 modulo p, and k_p must divide m, where n=(p^i)*(m) and p and m are relatively prime.

Also, any conjugate of a Sylow p-subgroup is again a Sylow p-subgroup. As a consequence, if there is only one Sylow p-subgroup, then it must be normal.

For example, 2002=2*7*11*13. Suppose, for example, p=7. Then m=286. The only numbers that are simultaneously congruent to 1 modulo 7 and divide 286 are 1 and 22. If you can rule out 22 as an option by some other argument, you're done. Another approach is to try some of the other primes that divide 2002.

Answer 3

If you have 2 7 11 and 13 in the factorization of the order, that should give you a hint. Under the definition of a simple group. Look in your text.

Answer 4

A group of order 2002 has 2003 terms. Simple as that!

...how is 2003 terms simple? Unless of course, it's (x+1)^2002.

Answer 5

wow this is first time i heard of abstract algebra.. never learned it tho

Answer 6

2{3012y7-903893}2a*(30932+2)