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Let $G$ be a group of order 20, Prove that $G$ has a normal subgroup of order 5.

Obviously by Sylow theorem there is a subgroup of order 5, and since all Sylow p-subgroups are conjugate the only problem is to show that there is only one sungroup of order 5.

any help would be appreciated

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    $\begingroup$ What does Sylow say about how many order-5 subgroups there can be? $\endgroup$ – Chappers May 27 '18 at 18:40
  • $\begingroup$ en.wikipedia.org/wiki/Sylow_theorems, take a look at theorem 3 $\endgroup$ – The Giraffe Guy May 27 '18 at 18:42
  • $\begingroup$ You already have the answer :). In fact this unique Sylow 5-subgroup is even a characteristic subgroup (something more than being normal). $\endgroup$ – Bumblebee May 27 '18 at 19:18
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$|G|=20=5\cdot 2^2$. Now, let $n_p(G)$ be the number of Sylow $p$-subgroups of $G$.

Then, Sylow III says

  • $n_5(G)\equiv 1\bmod 5$
  • $n_5(G)$ divides $4$

So, $n_5(G)=1$. This means there is only one Sylow $5$-subgroup of $G$; which in turn, is a normal subgroup of $G$.

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As you said Sylow's thoerem gives there is a 5-sylow subgroup.

The number of sylow 5 subgroups $n_5$ divides 4 and $n_5$ is congruent to 1 modulo 5.

So $n_5=1+5k$ for some $k$ and $n_5=1+5k$ divides $4$. Since $n_k$ is a natural number $k\ge 0$ but if $k>0$ then $n_5>4$ contradicting that $n_5$ divides 4. Thus $n_5=1$

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Or we can say just use the fact that there is only 1 group of order 5 which is cyclic group of order 5 and it is normal, we also know that all p subgroups are conjugate, but cyclic group of order 5 is normal so it only conjugates with itself.

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    $\begingroup$ A cyclic subgroup of order $5$ need not in general be normal (it just happens to be so here due to the order of the group being $20$). $\endgroup$ – Tobias Kildetoft May 28 '18 at 8:48
  • $\begingroup$ I am new to group theory, but I think all cyclic groups are abelian which is a stronger property of being normal. Correct me if I am wrong. $\endgroup$ – alper akyuz May 28 '18 at 8:58
  • $\begingroup$ math.stackexchange.com/questions/689577/…, for reference I think this is sufficent. $\endgroup$ – alper akyuz May 28 '18 at 9:00
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    $\begingroup$ That would be sufficient if you had argued that the condition given in the link holds here, which you have not. And no, being abelian and being normal are completely unrelated (you are probably thinking of the similarly sounding statement that all subgroups of abelian groups are normal). $\endgroup$ – Tobias Kildetoft May 28 '18 at 9:03
  • $\begingroup$ Thanks, I guess I have more to learn. $\endgroup$ – alper akyuz May 28 '18 at 9:19

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