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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 $H=\big<a: a^p=1\big>.$ 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 $H=ghg^{-1}$.

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

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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 $G$.

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    $\begingroup$ 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. $\endgroup$ – Tobias Kildetoft Mar 22 at 9:08
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    $\begingroup$ @ThomasShelby No, it is the subgroup generated by all elements of order $p$. $\endgroup$ – Tobias Kildetoft Mar 22 at 9:12
  • $\begingroup$ @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 ..."? $\endgroup$ – lesnik Mar 22 at 9:12
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    $\begingroup$ @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). $\endgroup$ – Tobias Kildetoft Mar 22 at 9:19
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    $\begingroup$ @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). $\endgroup$ – Tobias Kildetoft Mar 22 at 9:30

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