Question

If p is prime and G is a subgroup of ${\mathit{S}}_{\mathbf{p}}$that contains a transposition and a p-cycle, prove that . [Exercise 8 is the casep=5 .]

Short answer

It is proved that $G=S_p$.

Step-by-step solution

Determine splitting field

The shortest field extension of the extension field over which a minimum polynomial break or dissociates into linear components is known as the splitting field.

Determine the proof

Suppose a transposition is contained in a subgroup G. Now, suppose two elements$a,b\in G$ and this will have its own inverse.

Since, we have $\left(ab\right)\left(ab\right)=1$, and it is clear that splitting field F of $f\left(x\right)$ over $\mathrm{\mathbb{Q}}$. The splitting field consist a single pair of complex conjugate root represents an order 2 automorphism that was assumption. This means $G=Gal(F/\mathrm{\mathbb{Q}})$.

Now,$S_5$ is a subgroup of order 5, contains subgroup whose order is 2 which is generated by single 2 cycles. This means $[F:\mathrm{\mathbb{Q}}]=p$.

The Galois theory used to determine that the order of the groupG divided by p, and G contains ap cycle by using Cauchy’s Theorem.

Here,$S_p$ is generated by ap cycle and a 2 cycle, also a subgroupG is contained by $S_p$.

Now, from the above statements it is clear that ifG containp cycle and that cycle generates $S_p$, this means that subgroup is same as its group.

Thus,$G=S_p$ .