I don't understand the part of the followng proof that I underlined with red. I don't see why $PN/N \cong P/(P \cap N)$ implies $PN/N$ is a p-subgroup of $G/N$.
Asked
Active
Viewed 62 times
1 Answers
0
A quotient of a $p$-group is a $p$-group. This applies in the infinite case, where we define a $p$-group to be one all whose elements have prime power order $p^k$, for some $k$.
Pedro
- 122,002
-
You're not being clear enough in your answer, how does the conclusion follow from the two groups being isomprphic? Or are you saying we don't need that? We're talking about the finite case here, so the infinite case is not a problem. But what I don't understand is this: $|P|=p^k$ for some $k \ge 1$ and $|P \cap N|=p^r$ for some $r \ge 1$, but what if $k=r$? Then $|P|/|P \cap N|=1$ and $P/ P\cap N$ is not a p-subgroup (unless we consider $p^0$). – Twnk Apr 19 '14 at 23:40
-
@Twink $P$ is a $p$-group. $P/(P\cap N)$ is a quotient of a $p$-group, hence a $p$-group. Then $PN/N$ is a $p$-group. The trivial group is trivially $p$-group, in fact it is the only group that is a $p$ group for any prime $p$. – Pedro Apr 19 '14 at 23:42
-
So, in the definition of $p$-group, do we consider the power $p^0$ of $p$? – Twnk Apr 19 '14 at 23:45
-
Yes, but things are boring in that case, you get the trivial group. – Pedro Apr 19 '14 at 23:45
-
Ok but we need that to assure that the quotient of a $p$-group is a $p$-group, cause we could have $|P/P\cap N|=1$, that was my problem understanding this proof. Thanks. – Twnk Apr 19 '14 at 23:47