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  1. Prove that any subgroup of order $ p^{n-1} $ in a group $G$ of order $p^{n}$, p a prime number, is normal in $G$.

  2. $(a)$ Prove that a group of order 28 has a normal subgroup of order 7.

To deal with (a), can I just say 7 | 28, by Sylow's Theorem, there should be some Sylow 7-subgroups. Also, 7k+1 | 4 (=28/7), k=0. That means, there is a unique Sylow 7-subgroup.

(A Sylow k-subgroup = a subgroup of order k? In what situation this subgroup is normal?)

$(b)$ Prove that if a group $G$ of order 28 has a normal subgroup of order 4, then $G$ is abelian.

Note: Sorry that I am a student currently study "Sylow's Theorem", I don't really know any skill to deal with a group just with "the order". Would you mind to explain in detail?

Thanks a lot !

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  • $\begingroup$ Ad 1: Suppose $G$ is a group of order $p^n$, and $H$ is a subgroup of order $p^{n-1}$ of $G$. If $H$ were not normal, can you let $G$ act on the set of conjugates of $H$? $\endgroup$ – Daniel Fischer Dec 21 '14 at 13:18
  • $\begingroup$ I am sorry that I don't know the relation between normal and conjugate... If H were nor normal, should there be more than 1 conjugate class? $\endgroup$ – Richard Dec 21 '14 at 13:26
  • $\begingroup$ If $H$ were not normal, then $H$ would have more than one conjugate (itself), there would be a $g\in G$ with $gHg^{-1} \neq H$. The point of my comment is (a case of) the lemma in mesel's answer. Although, maybe it's better to let $G$ act on the set of cosets of $H$. $\endgroup$ – Daniel Fischer Dec 21 '14 at 13:28
  • $\begingroup$ Great! Thanks, you make me a clear picture of those relations! $\endgroup$ – Richard Dec 21 '14 at 13:35
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I will write a very useful lemma,

Lemma: If the index of $H$ in $G$ is the smallest prime dividing $G$, then $H$ is normal.

This lemma proves $1$ directly.

2) Notice that whhen you show that it has a uniqe Sylow-p subgroup, you also shows that it is normal in G.

$b)$ Let $H$ be the subgroup of order $7$ and $K$ be a subgroup of order $4$. It is clear that $H\cap K =1$ and both of them is normal. Then $HK\cong H\times K$. Since both $H,K$ is abelian then $H\times K$ is abelain. Thus, $HK=G$ is abelian.

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  • $\begingroup$ May I ask, how to prove this Lemma? I don't know if Sylow-p subgroup is unique, then it must be normal ? $\endgroup$ – Richard Dec 21 '14 at 13:27
  • $\begingroup$ @user131605 That has been asked somewhere on the site previously. I know I answered it, so I should be able to find it again, but it is a bit tricky from my phone. If nobody else links it and you cannot find it, I will try to remember to dig it up when I get to a computer. $\endgroup$ – Tobias Kildetoft Dec 21 '14 at 13:50
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    $\begingroup$ a Lemma's proof can be found here juanmarqz.wordpress.com/2013/10/18/… it uses the Double Coset Counting Formula. $\endgroup$ – janmarqz Dec 21 '14 at 14:09
  • $\begingroup$ I see you comment now. I guess your qustions was answered in other comments and answered. $\endgroup$ – mesel Dec 21 '14 at 14:49
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    $\begingroup$ @user131605: We took $K$ sylow-2 subgroup. And by defination it is the group of order $2^k$ where $2^k$ is the larget possible number dividing $|G|$. Thus, $|K|$ must be $4$. $\endgroup$ – mesel Dec 21 '14 at 19:57
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1) Consider the action of $G$ on the lateral classes of the subgroup ( call it H ) $$ g \cdot kH = gkH $$ This is an action. Consider the kernel K of the action, i.e $$K = \lbrace g \in H | g \cdot wH = w H \ \ \forall \ wH \rbrace$$Then $K \trianglelefteq G$; moreover $K \subseteq H $ ( simply consider $wH = H $ in the definition of $K$).

$G/K$ is isomorphic to a subgroup of $S_p$ because the lateral classes of $H$ are $p$ so $|G/K| \mid p!$ . But $|G/K|$ is a power of $p$ and this implies that $|G/K| = p$, otherwise $|G/K| $ can't divide $p!$ . Thus $|K| = |H| $ and $K \subseteq H \Longrightarrow K = H $

2) Let $n_7$ be the number of $7$-sylow subgroups. Then by Sylow theorems $$n_7 \equiv 1 \mod 7 $$ $$n_7 \mid 4 $$ Thus $n_7 = 1 $, there is an unique $7$-sylow and so it is normal. Infact suppose $H$ is the unique sylow and let $g \in G $. $$g^{-1}Hg $$ is another $7$-sylow because has the same cardinality, but there is only one sylow, so $$g^{-1}Hg = H$$This is true for all $g \in G$ and so $H$ is normal.

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  • $\begingroup$ Same here, I don't know why an unique Sylow-p subgroup => it is normal. $\endgroup$ – Richard Dec 21 '14 at 13:33
  • $\begingroup$ @user131605: I have edited $\endgroup$ – WLOG Dec 21 '14 at 13:37
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    $\begingroup$ @user131605, "Any two p-Sylow-Subgroups in a finite G are conjugated" is a Sylow's Thm. $\endgroup$ – janmarqz Dec 21 '14 at 15:01
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Another approach, using the rather important property that finite $\;p$- groups have non-trivial center, which also gives for free the existence of such groups (and even of a normal subgroup of order $\;p^k\;$ , for any $\;0\le k\le n\;$.

Take $\;1\neq z\in Z(G)\implies \langle z\rangle\lhd G\;$ , so

$$\left|G/\langle z\rangle\right|=p^{n-1}$$

Apply induction now (since the claim is clearly true for $\;n=1\;$ and even for $\;n=2\;$ ) and the correspondence theorem (CT): there exists a normal subgroup $\;\overline K\lhd G/\langle z\rangle\;$ of order $\;p^{n-2}\;$ , which by the CT corresponds to a normal subgroup $\;K\lhd G\;$ of order (by Lagrange) $\;p^{n-2}\cdot p= p^{n-1}\;$

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  • $\begingroup$ It should be made clear that this approach does not explain why all subgroups of index $p$ in $G$ are normal. $\endgroup$ – KCd Dec 21 '14 at 18:08

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