# Every normal Sylow $P$ subgroup of $G$ is fully invariant.

This particular result's proof is left as an exercise in textbook Thomas Hungerford and I couldn't prove it.

Result: Every normal Sylow p-subgroup of G is fully invariant.

Fully invariant: A subgroup H of a group G is called fully invariant if f(H) < H for every endomorphism f:G-> G.

Edit : This is also a result in proving fully invariant which is not proved in text.

If G is solvable and N is a minimal normal subgroup, then N is an abelian p-group for some prime p. Here in proof author wrote derived group of N is fully invariant in N. How?

I am not getting any thoughts on how endomorphism can be proved as f(H) <H.The problem occurs because of I am not able to understand how to use surjective property of endomorphism.

Any advice would be really appreciated.

• Hint: If a $p$-Sylow subgroup $P$ is normal then every element of order $p^{k}$ for some $k\in\mathbb{N}$ is in $P$. Now consider the order of images of elements in $P$. Aug 22, 2020 at 11:24
• @runway44 in the example you've given, $f(\mathbb{Z}_{2}\oplus 0)=0\oplus 0\le\mathbb{Z}_{2}\oplus 0$, so that's not a counter example. Aug 22, 2020 at 11:31
• Oops, I assumed $=$ and didn't even read $<$ in the definition of invariant. Aug 22, 2020 at 11:44
• Here is a hint for an elementary proof: Assume that there is some endomorphism $f$ for which is does not hold, and consider $f(P)P$.What is the order of this? Aug 22, 2020 at 13:29
• @MorA."If a p-Sylow subgroup P is normal then every element of order $p^{k}$ for some k∈N is in P". How to prove this statement? Also, can you please give a complete answer if you have some spare time? That would be really helpful to me. Sep 4, 2020 at 7:19

I'll assume we are only discussing finite groups for this question.
I'll also be using $$\mathbb{N}=\{0,1,2,...\}$$, that is $$\mathbb{N}$$ is the set of all non-negative integers.

First the lemma in my comment:

Lemma: Let $$G$$ be a finite group, let $$P$$ be a normal $$p$$-Sylow subgroup of $$G$$, and let $$S=\{g\in G\mid \text{exists }k\in\mathbb{N}\text{ such that }o(g)=p^{k}\}$$.
Then $$S=P$$

Proof: For all $$g\in S$$ there exists $$k\in \mathbb{N}$$ such that $$o(g)=p^{k}$$.
$$\langle g\rangle$$ is a $$p$$-subgroup of $$G$$ and therefore it's a subgroup of some $$p$$-Sylow subgroup of $$G$$, let that be $$P_g$$. By Sylow's theorems $$P$$ and $$P_g$$ are conjugate, namely there exists some $$h\in G$$ such that $$P_g = h^{-1}Ph = P$$, the second equality is due to the normality of $$P$$.
Therefore $$\langle g\rangle\subseteq P$$, so $$g\in P$$.
From the above we get $$S\subseteq P$$.
Left to the reader: show that $$P\subseteq S$$ and thus $$P=S$$.
$$\blacksquare$$

From here we'll prove the main statement in the question:

Let $$G$$ be a finite group, and let $$P$$ be a normal $$p$$-Sylow subgroup of $$G$$.
The $$P$$ is fully invariant.

Proof: Let $$f : G\to G$$ be a homomorphism.
For all $$g\in f(P)$$, there exists $$h\in P$$ such that $$g=f(h)$$.
From the lemma there exists $$k\in\mathbb{N}$$ such that $$o(h)=p^k$$.
Now $$g^{p^k}=f(h^{p^k})=f(1)=1$$, so $$o(g)\mid p^k$$, therefore there exists $$l\in\mathbb{N}$$ such that $$o(g)=p^l$$, and from the lemma we get that $$g\in P$$.
Therefore $$f(P)\subseteq P$$.
The above holds for all homomorphisms $$f : G\to G$$, hence $$P$$ is fully invariant.
$$\blacksquare$$

For the second question from the edit:

Here in proof author wrote derived group of N is fully invariant in N. How?

Here's a proof of the following statement

Let $$G$$ be a group (not necessarily finite), the derived group $$G'$$ is fully invariant.

Proof: Let $$S = \{[x,y]=x^{-1}y^{-1}xy\mid x,y\in G\}$$ be the set of all commutators in $$G$$, then by definition $$G'=\langle S\rangle$$.
Let $$f : G\to G$$ be a group homomorphism, first we will show that $$f(S)\subseteq G'$$:
For all $$g\in f(S)$$ there exist $$x,y\in G$$ such that $$g=f([x,y])$$, since $$f$$ is a homomorphism we get:
$$g=f([x,y])=f(x^{-1}y^{-1}xy)=f(x)^{-1}f(y)^{-1}f(x)f(y)=[f(x),f(y)]\in G'$$ Therefore $$f(S)\subseteq G'$$, and so $$f(G')=f(\langle S\rangle)=\langle f(S)\rangle\subseteq G'$$, or $$f(G')\subseteq G'$$.
The above holds for all homomorphisms $$f : G\to G$$, hence $$G'$$ is fully invariant.
$$\blacksquare$$
Left to the reader: Let $$G,H$$ be groups, let $$f : G\to H$$ be a homomorphism, then for any subset $$S\subseteq G$$, $$f(\langle S\rangle)=\langle f(S)\rangle$$.

• That's an excellent answer. I really appreciate the beautiful explanation. Thankyou!
– Esha
Jan 16 at 18:51