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The question is

Let $G$ be a group such that $|G| = pk$, where $p$ is a prime, $k < p$. Then $G$ contains a normal subgroup of order $p$.

It is easy to use Cauchy's Theorem to see that there exists a subgroup $H$ of $G$ with order $p$, but how to continue from here? Thanks.

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  • $\begingroup$ To add on, this is 1st year Group Theory, so Sylow's Theorem is not expected $\endgroup$
    – Bernoulli
    Jun 13, 2017 at 20:52
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    $\begingroup$ I saw sylow theorems in my first year group theory lecture though $\endgroup$
    – user370967
    Jun 13, 2017 at 20:59
  • $\begingroup$ @Bernoulli, I added a different proof. $\endgroup$ Jun 13, 2017 at 21:18
  • $\begingroup$ Thank you very much everybody. You are very helpful. $\endgroup$
    – Bernoulli
    Jun 15, 2017 at 20:24
  • $\begingroup$ math.stackexchange.com/questions/1932206/… you can also check this, I think it can be useful. $\endgroup$
    – MOVZBL
    Jun 17, 2017 at 16:58

3 Answers 3

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Let $n_p$ be the number of subgroups of order $p$. Then by Sylow's theorem $$n_p | k$$ and$$n_p\equiv1 \ mod \ p$$ As $k<p$, therefore $n_p =1$. Hence, the group is normal.

Edit You mentioned you don't know Sylow's so I'm adding a different proof. From future please mention what you know about the subject in the question itself.

Let $H$ be a subgroup of order $p$, $g \in G$ and $K=gHg^{-1}$. Then $|K|=p$. Now $$|HK|=\frac{|H||K|}{|H \cap K|}$$ If $|H \cap K|= 1$, then $|HK|=p^2$. But $|G| < p^2$, therefore $|H \cap K|\neq 1$. This forces $|H \cap K|= p$ which shows $H =K$. As $g \in G$ was arbitrary, therefore $H$ is normal.

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Don't know where you are in learning group theory, but I would consider the Sylow Theorems.

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suppose $H=<h>$ and $S=<s>$ are distinct subgroups of order $p$. consider the complex $HS$. since $$ |HS| \le pk \lt p^2 $$ there must be indices $a\ne b,c \ne d$ in $[0,p-1] \cap \mathbb{N}$ such that: $$ h^as^c = h^bs^d $$ but this gives $$ h^{a-b} = s^{c-d} $$ from which it follows (since $a-b$ is invertible in $\mathbb{Z}_p$) that $H$ and $S$ are the same subgroup, counter to our assumption.

since the subgroup of order $p$ guaranteed by Cauchy's theorem is unique, it is normal. can you show that it is in fact a characteristic subgroup?

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  • $\begingroup$ $HS$ has more elements then $G$ what is point of rest of the argument ? $\endgroup$
    – mesel
    Jun 14, 2017 at 2:17
  • $\begingroup$ you try to avoid using the fact $|HK|=p^2$ ? $\endgroup$
    – mesel
    Jun 14, 2017 at 2:18

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