00:02
Let g be a group and the order of the group is equal to 35 but this is equal to since 35 is 5 into 7 this is equal to 5 cube times 7 cube.
00:22
So by first point by sylow theorem there will exist a sylow 5 subgroup of g.
00:31
There will exist a sylow 5 subgroup of g of order 125 or 5 cube.
00:55
We will prove that this subgroup is normal.
00:58
Let the number let n5 be the number of 5 sylow subgroups of g.
01:05
Then by sylow theorems n5 must divide 7 cube and n5 is congruent to 1 modulo 5.
01:24
Now 7 cube is the power of a prime so its factors are 1, 7, 7 square and 7 cube.
01:36
These are all the factors of 7 cube.
01:39
So this set is equal to 1, 7, 49 and 7 cube is 343.
01:57
Now np has to be n5 has to be one of these numbers because yeah because it has to divide 7 cube it has to be one of the factors so it has to be one of these numbers and np is also congruent to 1 sorry n5 is also congruent to 1 modulo 5.
02:19
So it can be 1 it cannot be 7 because this is congruent to 2 modulo 5 this cannot be 49 also as its congruent to 4 modulo 5 and this cannot be 343 as its congruent to 3 modulo 5.
02:34
So the only possibility is n5 is equal to 1...
