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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 !

WLOG
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Richard
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  • 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 $H$? – Daniel Fischer Dec 21 '14 at 13:18
  • 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? – Richard Dec 21 '14 at 13:26
  • 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 $H$. – Daniel Fischer Dec 21 '14 at 13:28
  • Great! Thanks, you make me a clear picture of those relations! – Richard Dec 21 '14 at 13:35

3 Answers3

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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.

mesel
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  • May I ask, how to prove this Lemma?

    I don't know if Sylow-p subgroup is unique, then it must be normal ?

     – Richard Dec 21 '14 at 13:27
  • @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. – Tobias Kildetoft Dec 21 '14 at 13:50
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    a Lemma's proof can be found here http://juanmarqz.wordpress.com/2013/10/18/double-coset-counting-formula/ it uses the Double Coset Counting Formula. – janmarqz Dec 21 '14 at 14:09
  • I see you comment now. I guess your qustions was answered in other comments and answered. – mesel Dec 21 '14 at 14:49
  • @Mesel, may I ask how you prove the lemma? It is a little bit hard for me to understand the prove using double coset. – Richard Dec 21 '14 at 15:03
  • @user131605, here, WLOG did that – janmarqz Dec 21 '14 at 15:05
  • @user131605: if you are familiar with group action, Wlog provides a proof with group action. – mesel Dec 21 '14 at 15:09
  • Sorry that I found a question, a stupid one, why can't it be $H$ x $K$ x $P$ with order 7 x 2 x 2? – Richard Dec 21 '14 at 15:56
  • @user131605: I did not get what $P$ is. $|H|=7$ and $|K|=4$ and $$|HK|=\dfrac{|H||K|}{|H\cap K|}$$ And since $|H\cap K|=1$ then $|HK|=28$ which memans $HK=G$. – mesel Dec 21 '14 at 19:45
  • I just wonder if $|H|=7$, $|K|=2$, $|P|=2$, is a possible situation for $H, K, P$ are normal subgroups. – Richard Dec 21 '14 at 19:54
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    @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 $4$. – mesel Dec 21 '14 at 19:57
  • @user131605 The previous question is here http://math.stackexchange.com/questions/164244/normal-subgroup-of-prime-index/ and it has a couple of different proofs. They all use group actions to some extend, though some Places it is hidden (like in mine where it is hidden in the fact that groups of order $p^2$ are abelian). – Tobias Kildetoft Dec 22 '14 at 11:42
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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.

WLOG
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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) $\;p^{n-2}\cdot p= p^{n-1}\;$

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