# Show that $G$ is not solvable.

group-theory

Let $$p = 2k + 3$$ be a prime number $$\ge 5$$. Let $$G$$ be a subgroup of $$S_p$$ containing a $$p$$-cycle and an involution $$g$$ that fixes exactly three elements, i.e., a product of $$k$$ disjoint transpositions. Can I conclude that $$G$$ is not solvable? If so, how?

EDIT 1: If $$p = 5$$, then $$g$$ is a transposition, so $$G$$ is actually all of $$S_5$$, which is known not to be solvable. But what about the general case?

EDIT 2: Let's attack this problem from a different angle. Let $$g \in S_p$$ be an involution and let $$h \in S_p$$ be a $$p$$-cycle. Write $$G = \langle g, h \rangle$$ and $$H = \langle h \rangle$$.

Then the following propositions are equivalent:

• $$ghg$$ is a power of $$h$$.
• $$H$$ is a normal subgroup of $$G$$.
• $$H$$ is the only Sylow $$p$$-subgroup of $$G$$.

(I am still stuck here.)

Yes you can conclude that $$G$$ is not solvable, and many of your assumptions are unnecessary.
Let $$p$$ be prime, and let $$G$$ be a solvable subgroup of $$S_p$$ containing a $$p$$-cycle. Then $$G$$ is transitive and hence primitive, so its nontrivial normal subgroups are transitive.
Let $$P$$ be a minimal normal subgroup of $$G$$. Then $$P$$ is abelian and transitive, so $$|P|=p$$, and hence $$G$$ is contained in the normalizer of $$P$$ in $$S_p$$, which is a $$2$$-transitive group of order $$p(p-1)$$ in which no non-identity element fixes more than one point