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Let $K$ be a Sylow subgroup of a finite group $G$. Prove that if $x \in N(K)$ and the order of $x$ is a power of $p$, then $ x \in K$.

This is how I tried...

Since $K$ is normal in $N(K)$, the factor group $$\frac{N(K)}{K}=\{K,xK,....\} $$ is justified.

Then I took $$(xK)^{P^t}=x^{P^t}K=K$$.

This is where I am stuck.

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  • $\begingroup$ You have not said what $p$ is. $\endgroup$
    – Derek Holt
    May 12, 2016 at 8:09

5 Answers 5

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You can also do this in a more elementary way not using the quotient group.

As already noted $K$ is normal in $N(K)$. This implies that $K$ is the only $p$-Sylow-subgroup of $N(K)$ (since $N(K)$ certainly has no $p$-subgroups of larger order and all $p$-Sylow-subgroups of $N(K)$ are conjugate under $N(K)$). Hence the group generated by $x$ (a $p$-subgroup of $N(K)$) lies in $K$ (since it must lie in SOME $p$-Sylow-subgroup of $N(K)$ and there is only one).

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    $\begingroup$ Nice answer. Just one comment with the "more elementary way". Tquotient group is something that one can do directly after having defined what is a group, and the proof that all $p$-Sylow subgroups are conjugate use in fact the cosets, which is basically the same as looking at the quotient. $\endgroup$ Apr 10, 2014 at 12:57
  • $\begingroup$ You're certainly correct. "Elementary" was perhaps not the right choice of words. $\endgroup$ Apr 14, 2014 at 13:47
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By definition of $p$-Sylow, the group $G$ has order $p^m a$, where $a$ is coprime to $p$, and $p^m$ is the order of $K$.

As you said, $K$ is normal in $N(K)$, which is a subgroup of $G$. The order of $N(K)$ divides then the order of $G$ and is a multiple of the order of $K$. Hence, it has order $p^m b$ where $b$ divides $a$.

The order of $N(K)/K$ is then equal to $b$, which is coprime to $p$. The image of $x$ by the group homomorphism

$$N(K)\rightarrow N(K)/K$$ has an order which divides the order of $x$. Since by assumption the order of $x$ is $p$, then the image has an order which divides $p$. As $N(K)/K$ has order $b$, prime to $p$, this implies that $x$ is sent onto the identity, and then belongs to $K$.

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Here is the rough idea. See if you can fill in the details:

If $x\notin K$, then $\langle x\rangle K$ would be a proper supergroup of $K$ (this is where we needed the normalizing property) of order

$$|\langle x\rangle K|=\frac{|\langle x\rangle||K|}{|\langle x\rangle K|}\geqslant p|K|$$

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  • $\begingroup$ still not getting it,plz a elaborate a little bit.. $\endgroup$ Apr 10, 2014 at 10:34
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Since $K$ is normal in $N(K)$,consider the cyclic group generated by $x$,which is $<x>$,and you need to prove that $<x>K$ is a subgroup of $N(K)$,which implies that $<x>K$ is a subgroup of $G$,the group where $K$ is in.Then consider $|<x>K|=\frac{|<x>||K|}{|<x>\cap K|}$,or $\frac{|<x>K|}{|K|}=\frac{|<x>|}{|<x>\cap K|}$.$K$ is a Sylow-p subgroup of $G$ ,and $<x>K$ is a subgroup of $G$,thus $\frac{|<x>K|}{|K|}$ is relatively prime with $p$(we cannot find a subgroup which order is a power of $p$ and larger than $K$),if $<x>\cap K$ is a proper subgroup of $<x>$,then we will conclude that $\frac{|<x>K|}{|K|}$ has a factor $p$,this absurdity shows that $<x>\cap K=<x>$,and immediately $x\in K$.

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Clearly, $K$ is normal in $N(K)$ so we can talk about the factor group $\frac{N(K)}{K}$. Since $G$ is finite and K is a Sylow p-subgroup of $G$ so $\frac{|N(K)|}{|K|}$ doesn't have any power of $p$ and it is given that $|x|$ is a power of $p$ so $\gcd(|x|,\frac{|N(K)|}{|K|}) = 1$ $\implies \exists \ m,n\in \mathbb{Z}$ such that $|x|m + n\frac{|N(K)|}{|K|} = 1$. and since $x\in N(K)$ we know that $x^{\frac{|N(K)|}{|K|}} \in K$. Now $x = x^{|x|m + n\frac{|N(K)|}{|K|}} = (x^{|x|})^m (x^{\frac{|N(K)|}{|K|}})^n = (x^{\frac{|N(K)|}{|K|}})^n \in K$ as $x^{\frac{|N(K)|}{|K|}} \in K$

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