MHB Number of elements and subgroups

[kalish1]

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

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

Thanks.

 

[Deveno]

This is an interesting problem.

I think you have established that $G$ contains a non-trivial normal subgroup. Now let's consider the orbit-stabilizer theorem in light of this:

We have, for $x \in \{H_1,H_2,H_3,H_4,H_5,H_6\}$:

$|G| = |Gx|\ast|\text{Stab}(x)|$

Since this action is transitive (any sylow 5-subgroup is some conjugate of $H_1$, for example) we have $|Gx| = 6$, which means that $|\text{Stab}(x)| = 25$.

Since $H_i \subseteq \text{Stab}(H_i)$, it is clear that these sylow 5-subgroups are indeed their own stabilizers under the action.

Thus the kernel of the action is the intersection of all the stabilizers, which means that:

$|\text{ker}(\phi)| = 5$.

This tells us that $|\phi(G)| = 150/5 = 30$.

Now any group of order 30 has a normal sylow 5-subgroup, which means that the pre-image of that subgroup (under $\phi$) is a normal subgroup containing $\text{ker}(\phi)$. But such a pre-image is a group of order 25 in $G$, which contradicts the non-normality of any 5-sylow subgroup in $G$.

Hence $G$ does not exist, and the answer to (a) and (b) is: 0.