# Thread: Number of elements and subgroups

1. ## Number of elements and subgroups

I have a question that I feel I am going about in a roundabout way, and would like some help on. I am preparing for an exam.

Problem: Let G be a group with |G|=150. Let H be a non-normal subgroup in G with |H|=25.

(a) How many elements of order 5 does G have?

(b) How many elements of order 25 does G have?

My attempt:

G has 6 subgroups of order 25 because the number of subgroups of order 25 has to divide $\displaystyle$150/25=6$and$6 \cong 1 (\mod 5)$$. So there are 6 such subgroups:$$\displaystyle \{H_1,H_2,H_3,H_4,H_5,H_6\}$$. Let G act on this set by conjugation. The permutation representation of the group action of conjugation on this set is \displaystyle \phi:G \rightarrow S_6, and |S_6|=720. Also, |\ker(\phi)||im(\phi)|=150$$. Because $\displaystyle$|im(\phi)|$$has at most one factor of 5, 5 divides$$\displaystyle |\ker(\phi)|$$. Suppose 25 divides$$\displaystyle |\ker(\phi)|$$. Then$$\displaystyle \ker(\phi)$$has a Sylow 5-subgroup H_i. So$$\displaystyle H_i \subset \ker(\phi) \implies aH_ia^{-1} \subset a(\ker(\phi))a^{-1} = \ker(\phi)$$. By the Second Sylow Theorem,$$\displaystyle H_1 \cup \ldots \cup H_6 \subset \ker(\phi) \implies G = \ker(\phi) \implies \phi$trivial$. So $\ker(\phi)$ has a subgroup $K$ with $|K|=5$. So for some $$\displaystyle H_i, \ker(\phi) \cap H_i=K \implies a(\ker(\phi))a^{-1} \cap aH_ia^{-1} = aKa^{-1}$$

What next? Surely this shouldn't be so long-winded.

2. ## Re: Number of elements and subgroups

Hi,
How's this for an answer? If G is of order 150 with a non-normal subgroup of 25, then G has 237 elements of order 5!! This is a true statement because there is no group of order 150 with a non-normal subgroup of order 25.

Let G be a group of order 150. Then the Sylow 5 subgroup of G is normal.
Proof.
The order of G is 75*2. So G has a normal subgroup N of order 75 -- see Huppert Endliche Gruppen I, page 30 or prove it yourself.
If you don't read German, I could supply the proof if necessary.
Thus by Sylow, N has a normal Sylow 5 subgroup which is then normal in G.

By the way, everything you say is true. This is typically the way you analyze groups of small order.

3. ## Re: Number of elements and subgroups

Thanks for the reply. I am not able to find the book online, so if you could supply the proof it would be great. I have a lot of other material to study for this exam anyway.

By the way, I understand that G has a normal subgroup N of order 75 because every subgroup of index 2 is normal.

4. ## Re: Number of elements and subgroups

Hi,
I hope the attachment answers all your questions. By the way, you're right that any subgroup of index 2 in a group is normal. But there need not be any subgroup of index 2. Example: A5 has order 60, but has no subgroup of order 30 since A5 is simple.

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### let g be a group of order75 what is the number of subgroup of order 25

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