Question

Question: If f is an automorphism of G and K is a Sylow p-subgroup of G, is it true that $\mathit{f}\mathbf{\left(}\mathit{K}\mathbf{\right)}\mathbf{=}\mathit{K}$ ?

Short answer

It is proved that$f\left(K\right)\ne K$

Step-by-step solution

Definition of homomorphism

An isomorphism from a group G onto itself is called an automorphism of G.

Given that f is an automorphism of G and K is a Sylow p-subgroup of G.

Proving that f(K)≠K

This can be proved by considering the given example:

Let $G=S_4$ and$S_4$ has order 24.

Then, $G=\left\{\left(1\right),\left(1234\right),\left(13\right)\left(14\right),\left(1432\right),\left(24\right),\left(12\right)\left(34\right),\left(13\right),\left(14\right)\left(32\right)\right\}$and $S^4$ has order $2^3.3$.

So, any subgroup of order 3 is a Sylow 3-subgroup.

Therefore, Sylow 3-subgroups can be written as:

$K=\left\{1,\left(132\right),\left(123\right)\right\}$

Suppose f is a conjugation by$\left(12\right)$ .

Find $f\left(K\right)$ as:

$$\begin{gathered} f\left(k\right)=\left\{1,\left(23\right),\left(13\right)\right\} \\ \ne K \end{gathered}$$

Hence, it is proven that$f\left(K\right)\ne K$.