Intereting Posts

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Combinatorics/Task Dependency
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Isometry from Banach Space to a Normed linear space maps
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Find the matrix of $\langle f, g \rangle = \displaystyle\int_0^1 f(t) g(t) \, \mathrm dt$ with respect to the basis $\{1,t,\dots,t^n\}$
An example showing that van der Waerden's theorem is not true for infinite arithmetic progressions
Relation between Hermite polynomials and Brownian motion (on martingale property)
Geometric construction of hyperbolic trigonometric functions
Does the limit of a descending sequence of connected sets still connected?
Making Tychonoff Corkscrew in Counterexamples in Topology rigorous?
Quadratic extensions of $\mathbb Q$

Let $p$ be a prime number and $G$ a finite group where $|G|=p^n$, $n \in \mathbb{Z_+}$. Show that any subgroup of index $p$ in it is normal in $G$. Conclude that any group of order $p^2$ have a normal subgroup of order $p$, but without using the Sylow theorems.

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- On solvable quintics and septics
- Homomorphism from $\mathbb{Q}/ \mathbb{Z}\rightarrow \mathbb{Z}/n\mathbb{Z}$
- Calculating the group co-homology of the symmetric group $S_3$ with integer coefficients.
- Ways $S_3$ can act on a set of 4 elements.
- Inverse Image of Maximal Ideals
- Constructing an example s.t. $\operatorname{Hom}_R(M,N)$ is not finitely generated
- Why is a ring $R$ with the property that $r=r^2$ for each $r\in R$ so special?

I think another approach in light of Don’s answer can be:

Lemma:Let $G$ is a $p$-group and $H<G$ then $H\lneqq N_G(H)$.

Here we know that $[G:H]=p$ then $H$ is a proper subgroup of $G$. So the lemma tells us in this group we have $H$ as a proper subgroup of its normalizer in $G$. In fact our conditions make $N_G(H)$ to be $G$ itself and this means that $H\vartriangleleft G$.

You only need the following

**Lemma:**: If $\,G\,$ is a finite group and $\,p\,$ is the smallest prime diving $\,|G|\,$ , then any subgroup of *index* $\,p\,$ is normal in $\,G\,$ .

**Proof (highlights):** Let $\,N\leq G\,\,,\,\,[G:N]=m\,$ , and define an action of $\,G\,$ on the set of $\,X\,$ of left cosets of $\,N\,$ by $\,g\cdot(xN):=(gx)N\,$ :

1) Check the above indeed is a group action on that set

2) Check that the given action induces a homomorphism $\,\phi:G\to S_X\cong S_m\,$ with kernel

$$\ker\phi=\bigcap_{x\in G}N^x\,,\,\,\,N^x:=x^{-1}Nx $$

(the above kernel is also called *the core* of $\,N\,$)

3) Check that $\,\ker\phi\,$ is the greatest subgroup of $\,G\,$ normal in $\,G\,$ which is contained in $\,N\,$

4) Now apply the above to the case $\,m=p=\,$ the smallest prime dividing the order of the group.

One more solution. This one I saw in an old paper (1895) by Frobenius (from here).

We proceed by induction. The case $n = 1$ is clear. Let $H$ be a subgroup of index $p$, ie. $H$ has order $p^{n-1}$. Since $p$-groups have nontrivial center, there exists $x \in Z(G)$ of order $p$. If $x \in H$, then $H/\langle x\rangle \trianglelefteq G/\langle x\rangle$ by induction and thus $H \trianglelefteq G$. If $x \not\in H$, then $G = H\langle x \rangle$ and $H \trianglelefteq G$ since $x$ is central.

This is a bit ad-hoc, but I thought up one more elementary solution.

Let $H$ be a subgroup of index $p$. Suppose that $H$ is not normal. Then there exists $g \in G$ such that $g^{-1}Hg \neq H$. Thus $G$ is equal to the product $H(g^{-1}Hg)$, but this is in contradiction with the following fact.

If $M$ is a subgroup of $G$ and $g \in G$ such that $G = Mg^{-1}Mg$, then $M = G$.

Proof: Since $g \in Mg^{-1}Mg$, we get $g \in M$ and thus $G = MM = M$.

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