# Cyclic sylow subgroup of order 9 with 3-core of order 3

Can a finite group G have a cyclic Sylow 3-subgroup of order 9, such that the intersection of the Sylow 3-subgroups of G has order exactly 3, without having non-identity normal subgroups of order coprime to 3?

In classifying the finite groups with cyclic Sylow 3-subgroups of order 9, it seems reasonable to split into cases based on the size of the 3-core (the intersection of the Sylow 3-subgroups) mod the 3'-core (the largest normal subgroup of order coprime to 3). 3-Cores of sizes 1 and 9 are easy to handle, but size 3 is devolving into a number of cases none of which seem to work, but for no systematic reason.

Is there some systematic reason it cannot occur, or is there an example I've overlooked?

Edit: The original left out the important condition on normal subgroups of order coprime to 3, without which the classification is intractable. Alex's answer shows how to use such normal subgroups to get fairly arbitrary behavior (for any type of Sylow).

Edit 2: If the perfect residuum X of the group is nontrivial, it contains Ω(P) of the cyclic Sylow p-subgroup P, and so Op(G) ≤ Op(X). In this case, Op(X) = 1. Hence the group is solvable, and Fit(G) = Op(G), so G/Fit(G) ≤ Aut(p) = p−1 has the wrong order.

• To be honest, you should have posted this as a separate answer and accepted it, since I don't think I contributed anything to it :-) – Alex B. Nov 26 '10 at 18:08
• Well, your contributions were helpful to me. And it strikes me as rude to blame an answer when the question was at fault. :-) – Jack Schmidt Nov 26 '10 at 18:22

Such groups are easy to construct: let $C_9$ act on something through a $C_3$-quotient. E.g. construct the semi-direct product $C_7\rtimes C_9$ such that $C_9$ acts on $C_7$ through the unique automorphism of order 3. Then, the Sylow 3-subgroups are not normal, so there is more than one of them, but there is a unique $C_3$, which is the centre of $G$.