# Group of order 21 and subgroup of order 7

I have this problem and I don't know how to solve it, because in this part of the book I only have normality and the isomorphism theorems.

Let $G$ a group and $|G|=21$ assume that $a \in G$ and $|a|=7.$ Prove that $A = \langle a \rangle$, the subgroup generated by $a$, is normal in $G$. i.e $$A \lhd G$$

At this moment of the book I CAN'T use Sylow theorems so I dont know how to do this without it. Thanks.

• My guess is to assume $A$ is not normal, thus there must be another subgroup of order $7$, then look at how $a$ acts on these subgroups by conjugation. Just a hunch, it's too late for any deep thinking from me. Aug 20, 2017 at 3:36

Suppose that $H$ and $K$ are distinct subgroups of $G$ with $|H| = |K| = 7$. Then, since $7$ is prime, we must have $H \cap K = 1$, and so $$|HK| = \frac{|H||K|}{|H \cap K|} = \frac{49}{|H \cap K|} = 49$$ This obviously can't happen in a group containing only $21$ elements. Therefore $\langle a \rangle$ must be the only subgroup of order $7$.

This implies that $\langle a \rangle$ is normal (indeed, characteristic).

Proof: any automorphism $\phi : G \to G$ must map $\langle a \rangle$ to a subgroup of the same size, hence must map $\langle a \rangle$ to itself. Conjugation by $g$ is an automorphism, hence $g\langle a \rangle g^{-1} = \langle a \rangle$ for all $g \in G$.

• Ah, nice. Having flashbacks to my algebra qualifiers now.... Aug 20, 2017 at 4:26
• Note that the same proof works for any $|G| \leq 48$. When $|G| = 49$ we have a counterexample: $G = \mathbb Z_7 \times \mathbb Z_7$ has eight subgroups of order $7$.
– user169852
Aug 20, 2017 at 4:39

$A$ is of index $3$, the smallest prime dividing $|G|$. Thus $A$ is normal. See Normal subgroup of prime index

• I can't use that, that is from 2 chapters ahead... I am looking for an "easier" way to solve it without using stronger theorems Aug 20, 2017 at 3:23