Two problems in Group theorem related to Sylow's theorem(maybe)

Question

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 !

Answer 1

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.

Answer 2

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.

Answer 3

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