Let G be a group of order 715=5 ·11 ·13 . Let H be a Sylow 13-subgroup of G and K be a Sylow 11

step 1 So, if x1, x2 belongs to k, then x1, x2, square equal x1 square, x2, belonging to h. Thus, x1, x

Let G be a group of order 715=5 ·11 ·13 . Let H be a Sylow...

Let $G$ be a group of order $715=5 \cdot 11 \cdot 13 .$ Let $H$ be a Sylow 13-subgroup of $G$ and $K$ be a Sylow 11 -subgroup of $G$. Prove that $H$ is contained in $Z(G)$. Can the argument you used to prove that $H$ is contained in $Z(G)$ also be used to show that $K$ is contained in $Z(G) ?$

Key Concepts

Sylow Theorems

The Sylow theorems give conditions on the number and structure of subgroups whose orders are powers of primes dividing the group order. In this context, they are used to determine that a Sylow subgroup might be unique (and hence normal) in the group. The uniqueness of a Sylow subgroup has important implications on how it behaves under conjugation by other elements of the group.

Conjugation Action and Centralizers

The conjugation action of a group on one of its subgroups leads naturally to a homomorphism from the group into the automorphism group of the subgroup. The kernel of this action is precisely the centralizer of that subgroup – the set of elements in the group that commute with every element of the subgroup. If the image of the conjugation map is trivial, then every element of the Sylow subgroup commutes with every element of the group, placing it in the center.

Automorphism Group of a Cyclic Group

When a Sylow subgroup is cyclic, its automorphism group is typically well-understood and is a group of units modulo the order of the subgroup. This structure allows one to constrain the possible sizes of the image of the conjugation homomorphism. In particular, if the order of the ambient group has no non?trivial common factors with the order of the automorphism group, then the conjugation action must be trivial, forcing the subgroup into the center.

Center of a Group

The center of a group consists of all elements that commute with every element of the group. A subgroup contained within the center is said to be central. In problems like this, one often shows that a subgroup is central by proving that the conjugation action is trivial, meaning every element of the subgroup remains fixed under conjugation by any group element.

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