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If a group is of order $20$. Its factorisation is $2^2*5$. So, there are subgroups of order $2,4$ and $5$. Whether a $2$-Sylow subgroup means subgroup of order $2$ or $4$ ($2^2$) ??

From Sylow's third theorem i can say there is a normal subgroup of order $5$ ($1+5k|20$) and number of $2$-sylow subgroups are ($1+2k|20$) = $2$ . So there are $2$ subgroups order $4$. Correct me, if i am wrong.

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  • $\begingroup$ You can say what you want when you define a name for the sylow groups. Some authors say $p$-Sylow group and some authors say $p$-Sylow subgroup when they refer to the subgroups of prime power of highest order of $p$. Yes, there is exactly one $5$-Sylow group, which is normal. Next, the number of $2$-Sylow groups must divide $5$ and is of the form $1+2k$, hence the number of $2$-Sylow groups is $1$ or $5$. $\endgroup$ – Fakemistake Jan 6 '18 at 19:05
  • $\begingroup$ I have always heard p-group for any subgroup having an order of a power of p and p-sylow subgroup for a p-group having maximal order. The 2nd sylow theorem then says every p-group is contained in a p-sylow subgroup. $\endgroup$ – David Reed Jan 7 '18 at 1:39
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Sylow-p subgroups are defined to be subgroups of size the largest power of $p$ which divides the order of the group.

As for your workings, not quite. It is true that there is a subgroup of order $2$ by Cauchy's theorem, but you are trying to apply the Sylow theorems to a subgroup which is not a Sylow $p$ subgroup.

It is true that there is one Sylow-5 subgroup and it is indeed normal (why?).

You have that there is either $1$ or $5$ Sylow $4$ subgroups since $$ n_4=1+2k $$ and $$ n_4\vert 5 $$

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  • $\begingroup$ So, i can find number of subgroups of order $4$ , which is obtained by $1+4k$ . Can i say anything about number of subgroups of order $2$ (which is not a Sylow-p subgroup) ?? $\endgroup$ – Bharadwaj Rcr Jan 6 '18 at 19:12
  • $\begingroup$ $4$ is a prime? $\endgroup$ – Fakemistake Jan 6 '18 at 19:18
  • $\begingroup$ ha you're right, just a difference in terminology then $\endgroup$ – qbert Jan 6 '18 at 19:19
  • $\begingroup$ So the conclusion is : a)The number of subgroups of order $4$ is $1$ or $5$ which we can calculate by $1+2k$ . Right ? b) We can only say there is a subgroup of order $2$ and cannot say anything about number of subroups of order $2$. $\endgroup$ – Bharadwaj Rcr Jan 6 '18 at 19:37
  • $\begingroup$ you can use the sylow theorems and the semidirect product to classify all groups of order 20, so you should, with a little work, be able to say a lot about the number of subgroups of order 2 $\endgroup$ – qbert Jan 6 '18 at 19:40

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