Question

If H is a proper subgroup of G, prove that G is not the union of all the conjugates of H. [Hint: Remenber that H is a normal subgroup of N(H); Theorem 9.25 may be helpful.]

Short answer

It is proved that, G is not the union of all the conjugates of H.

Step-by-step solution

Given information

It is given that H is a proper subgroup of G.

Also, we know that H is a normal subgroup of N(H).

Use Lagrange's Theorem

Let there be$\left[\mathrm{G}:\mathrm{N}\left(\mathrm{H}\right)\right]=\mathrm{n}$ distinct conjugates of H in G.

Also, by the Lagrange theorem$\mathrm{n}.\left|\mathrm{H}\right|\le \mathrm{n}.\left|\mathrm{N}\left(\mathrm{H}\right)=\mathrm{G}\right|$

Prove that n=1

Any two of these conjugates have at least $\left\{e\right\}$in common, possibly more.

Therefore, the number of elements in the union of all the conjugates of H is at most as:

$\begin{array}{rcl}1+\mathrm{n}.\left(\left|\mathrm{H}\right|-1\right)& =& \mathrm{n}-\left|\mathrm{H}\right|-\left(\mathrm{n}-1\right)\\ & \le & \left|\mathrm{G}\right|-\left(\mathrm{n}-1\right)\\ & \le & \left|\mathrm{G}\right|\end{array}$

If this union is all of G, these inequalities are equalities.

This implies, n=1.

Prove that G is not a union of conjugates of H

Since $\mathrm{G}=\mathrm{N}\left(\mathrm{H}\right)$and H is normal in G.

Then, the only conjugate of H is H itself, and $\mathrm{G}=$the union of the conjugate$=\mathrm{H}$

But H is given to be a proper subgroup.

So, G cannot be the union of the conjugates of H.

Therefore, G is not the union of all the conjugates of H.