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This is from Wikipedia Sylow's Theorem 2

Given a finite group $G$ and a prime number $p$, all Sylow p-subgroups of $G$ are conjugate to each other, i.e. if $H$ and $K$ are Sylow p-subgroups of G, then there exists an element g in G with $g^{−1}Hg$ = $K$.

My Question is

First question)

Since $H$ and $K$ are conjugate, If we defined the inner automorphism $\phi: H \to K $

Then Does it always mean H and K are isomorphic each other?

Though $H$ and $K$ would be different, Regarding the subgroups which is isomorphic as a same, Then Does it have to be one. But By the sylow third theorem, It looks like only true When the case of the normal group. I'm really confused between two theorem.

Second question)

Here is the question make me confused.

Find the subgroup of the $G$ $s.t. |G| =99$

By sylow third thm, The number of the sylow-3group, $P_3 =1$

So $P_3$ is the normal of the G meaning the $P_3$ subgroup is a unique.

So there are only one $P_3$ subgroup(This is just my thought).

But The one who suggest this question said there are two subgroups isomorphic with $Z_3 \times Z_3$ and $Z_9 $ respectively.

But Is it contradict with the $n_3=1$? Because the one is cylic the other isn't.

($n_3$ is the number of the sylow 3 subgroups.)

I'm really confused. Please help me and tell me What I was wrong. :(

Thank you.

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  • 3
    $\begingroup$ Regarding the first question, $H$ and $K$ need not be the same subgroup. "Isomorphic" means "same structure", not necessarily "same subgroup." You can think of them as being "copies" (maybe distinct, maybe not) of the same type of subgroup. Consider $S_3$. This group has three subgroups of order $2$. They are all cyclic with order $2$, hence all isomorphic. But they are three distinct subgroups. $\endgroup$ – Bungo Jun 25 at 6:18
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Yes, different Sylow-$p$ subgroups (of the same base group, for the same $p$) are necessarily isomorphic.

There is only one Sylow-3 subgroup in this example, but you can't tell whether it's $\Bbb Z_9$ or $\Bbb Z_3^2$ just from the order of $G$. It could be either. There is no reason why Sylow-$p$ subgroups should be abelian either, it's just that all groups of order $9$ happen to be abelian (this is true for the square of any prime).

And yes, if the number of Sylow-$p$ subgroups for some $p$ is $1$, then it is unique and there is only one (why are you stating the same thing in so many different ways?). The fact that it is normal follows from there, not the other way around.

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