Question

Let H be a subgroup of order n in a group G. If H is the only subgroup of order n, prove that H is normal. [Hint:Theorem 8.11 and Exercise 20 in section 7.4 ]

Short answer

It is proved that H is a normal subgroup.

Step-by-step solution

Important Theorem

Theorem 8.11: The following conditions on a subgroup of a group are equivalent:

1. Nis a normal subgroup of G.
2. $a^{-1}Na\subseteq N$ for every$a\in G$, Where $a^{-1}Na\subseteq N$$\left\{a^{-1}Na| n\in N\right\}$
3. $aN{a}^{-1}\subseteq N$for every $a\in G$, Where $aN{a}^{-1}\subseteq N$$\left\{aN{a}^{-1}| n\in N\right\}$
4. $a^{-1}Na\subseteq N$for every $a\in G$ .
5. $aN{a}^{-1}\subseteq N$ for every $a\in G$ .
   From Exercise 20, we know that$aN{a}^{-1}\in \left\{aN{a}^{-1}| n\in N\right\}$is a subgroup of G.

Proving that H is a normal subgroup

It is given that H is a subgroup of G and the order of His n.

Therefore, $\left|H\right|=n$

Now from exercise 20, we know that $a\in G$ and

$aH{a}^{-1}$ is also a subgroup of G

Since we know that conjugate subgroups must have the same order,

$\left|aH{a}^{-1}\right| = \left|H\right|$

Since H is the only subgroup of G with order n. Therefore, all the subgroups of G of order n must belong to H.

So, $aH{a}^{-1}\in H$

This implies .$aH{a}^{-1}\subseteq H$

Hence, from the above result and theorem 8.11, we can conclude that H is a normal subgroup.