# How to show that $H$ is normal in $G$?

Let $$G$$ be a group of order $$pq$$, where $$p$$ and $$q$$ are primes and $$p>q$$. Ler $$a\in G$$ be of order $$p$$ and $$H=\big.$$ Then $$H$$ is normal in $$G$$. I know that $$H$$ will be normal subgroup of G if $$gH=Hg$$ or $$H=gHg^{-1}$$. I tried as:

Let $$g\in G$$ and $$h\in H$$ then $$h$$ can be written in some power of $$a$$ but I don't, how to write $$g$$ and proceed to show that $$H=ghg^{-1}$$.

• Are you familiar with Sylow's theorem? – Thomas Shelby Mar 22 at 8:56
• It suffices to show $g^{-1}Hg\subseteq H$ for all $g\in G$. – Shaun Mar 22 at 9:00

The proof above very nearly works, so let's finish it up.

Lemma: If $$x^p=1, \text{ then } \forall g \in G (g^{-1}xg)^p=1,$$

Proof: Since consecutive pairs $$g^{-1}g=1$$ cancel each other, we have $$(g^{-1}xg)^p=g^{-1}x^pg=g^{-1}g=1.$$

Now assume $$a \in H$$. Then $$\exists x_1, x_2, \ldots x_n \in H \text{ such that } x_k^p=1 \text{ and }a=x_1x_2\cdots x_n$$. Then $$g^{-1}ag= \prod_{k=1}^n g^{-1}x_kg.$$

By the Lemma, each $$g^{-1}x_kg$$ is a generator of $$H$$ because it has order $$p$$, so $$g^{-1}ag \in H$$ and since $$a \in H, g \in G$$ were arbitrary, that shows $$\forall g \in G~g^{-1}Hg \subseteq H$$ so $$H \triangleleft G$$ and we are done.

Let's take any $$a$$ from $$H$$. Let's take some other element $$g$$ from $$G$$ and examine what is $$x = gag^{-1}$$. But

$$x^p = g a g^{-1} g a g^{-1}...g a g^{-1} = g a^p g^{-1} = g e g^{-1} = e$$

So, $$x$$ also belongs to $$H$$.

Conjugating by any element of the group $$G$$ leaves any element of $$H$$ inside $$H$$. That's actually a definition of normal subgroup. So, $$H$$ is normal in $$G$$.

UPDATE

As Tobias Kildetoft pointed out, this is not a proof yet. It is only a proof that conjugating leaves the set of all elements $$S = \{a: a^p=1\}$$ in place. The subgroup generated by these elements is a different thing!

Let's take some element $$h$$ of $$H$$. It is generated by elements of $$S$$: $$h = s_1 s_2... s_n$$

It's conjugated element: $$ghg^{-1} = gs_1 s_2... s_ng^{-1} = gs_1g^{-1} gs_2g^{-1}... gs_ng^{-1}$$

Is generated by $$gs_ig^{-1}$$. So, it is generated by by the elements of the same set $$S$$. So, it also belongs to $$H$$, which means $$H$$ is normal in $$G$$.

• The idea is correct, but there is a small detail missing: $H$ does not necessarily just consist of elements of order dividing $p$, but is generated by those. – Tobias Kildetoft Mar 22 at 9:08
• @ThomasShelby No, it is the subgroup generated by all elements of order $p$. – Tobias Kildetoft Mar 22 at 9:12
• @ThomasShelby I guess clarification is needed from the author. $H=\big<a: a^p=1\big>.$ - doesn't it read as "all $a$ such that ..."? – lesnik Mar 22 at 9:12
• @ThomasShelby Until you have shown that the subgroup in question is in fact of order $p$, you can't use Sylow. And the conclusion that it is of order $p$ is the same as there being just one Sylow $p$-subgroup (the subgroup in question is always normal, regardless of the group, it just happens to be of a nice form in this case). – Tobias Kildetoft Mar 22 at 9:19
• @ThomasShelby Sure, but that is not the subgroup we care about here (except that it turns out to be the same due to the order of the group). – Tobias Kildetoft Mar 22 at 9:30