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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.

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    $\begingroup$ 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 $P$. $\endgroup$
    – Mor A.
    Aug 22, 2020 at 11:24
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    $\begingroup$ @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. $\endgroup$
    – Mor A.
    Aug 22, 2020 at 11:31
  • $\begingroup$ Oops, I assumed $=$ and didn't even read $<$ in the definition of invariant. $\endgroup$
    – runway44
    Aug 22, 2020 at 11:44
  • $\begingroup$ 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? $\endgroup$ Aug 22, 2020 at 13:29
  • $\begingroup$ @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. $\endgroup$
    – Avenger
    Sep 4, 2020 at 7:19

1 Answer 1

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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 $S\subseteq G$, $f(\langle S\rangle)=\langle f(S)\rangle$.

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

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