Question

How many Sylow p-subgroup can G possibly have when

P= 3 and $\mathbf{\left|}\mathbf{G}\mathbf{\right|}\mathbf{=}\mathbf{72}$

Short answer

The Sylow 3-subgroups are $\{1,4\}$.

Step-by-step solution

Step-by-Step Solution Step 1: Third Sylow Theorem

The number of Sylow p-subgroups of finite group G divides $\mathbf{\left|}\mathbf{G}\mathbf{\right|}$ and is of the form 1+pk for some non-negative integer k.

Given that, p=3 and$\left|G\right|$ =72.

Finding Sylow 3-subgroups

By Third Sylow Theorem, the number of Sylow 3-subgroups of finite group G, which divides 72 and is the form of 1+3k can be found by using different non-negative values of k.

• Using k as 0 we get 1, which also divides 72.
• Using k as 1 we get 4, which also divides 72.
• Any further values of k will give a result, which will not divide 72.

Therefore, the subgroups are 1 or 4.

Hence, the possible number of Sylow 3-subgroups are $\{1,4\}$.