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My class is studying on Sylow $p$-subgroups, and I had been stuck for several hours on determining the order of a Sylow $p$-subgroup of a group $G$ of finite order.

I asked a previous question like this here on this site but I completely misinterpreted my own question (by the way, this was my previous question: Determine order of a Sylow p-subgroup)

Now, I'd put up a bounty on that question as soon as the site allows me to (in $2$ days), but my quiz on this Sylow stuff is in $2$ days. Therefore, can I please try again with this question here? I'm going to try to make it easy for myself and make up a group order, say $|G| =12=2^2 \cdot 3$.

So let $G$ be a group of order $12$. For each prime $p$ dividing $|G|$, determine the order of a Sylow $p$-subgroup.

Now, I'd show my work, but I would not know the very first correct step (hence, my failed attempt at my previous question). All my "work" is shown there via the above link. Therefore any starting hints are nice.

Community
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Cookie
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    Aren't you looking for the numbers of a sylow $p-$ subgroup for a prime dividing $|G|$ ?! – Mikasa Mar 03 '14 at 07:35
  • That's what I did (or at least tried to do) at my previous question. But here I'm looking for order, or size of a Sylow p-subgroup. – Cookie Mar 03 '14 at 07:37
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    If so; the answer which @ajd posted is enough for us. (-: – Mikasa Mar 03 '14 at 07:38
  • Hint: you already wrote the orders, you just did not separate them by a comma. They occur in the formula starting "$|G|={}$". – Marc van Leeuwen Mar 03 '14 at 07:39
  • I did look through my textbook and reread the whole Section 8.3 of Hungerford, 2nd edition, which covers Sylow theory. It takes forever for me to even remotely understand this stuff so that's why it took quite a few hours ... – Cookie Mar 03 '14 at 07:44
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    OK, I guess it must say somewhere that for any prime $p$ there is a subgroup of $G$ whose order is the highest power $p^k$ that divides $|G|$, and this is a Sylow $p$-subgroup of $G$. That is all you need to know for this question. – Marc van Leeuwen Mar 03 '14 at 07:48

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A Sylow $p$-subgroup is simply a subgroup whose order is the highest power of $p$ dividing the order of the group. So in your example, a Sylow $2$-subgroup would have order $4$ and a Sylow $3$-subgroup would have order $3$.

ajd
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  • Okay, I understand the highest power of $p^k$ dividing the order part (as it is the definition of Sylow $p^k$-subgroup, where $k$ is an integer, to begin with). So let's say I have a different finite group $G$ with order $260 = 2^2 \cdot 5 \cdot 13$, then a Sylow $2$-subgroup has order $4$, a Sylow $5$-subgroup would have order $5$, and a Sylow $13$-subgroup would have order $13$. – Cookie Mar 03 '14 at 07:40
  • @glacier: That is correct. – Mikko Korhonen Mar 03 '14 at 07:44