# P primary group and the correspondence theorem

1. Jun 17, 2012

### jz2012

Hi,

I have a question from "A first course in abstract algebra" by J. Rotman,

Hi, this is a question from " A first course in abstract algebra" by J. Rotman
define d(G) = dim(G/pG)

chapter 5, lemma 5.8 (P392),

Let G be a finite p primary abelian group.
If S<=G, then d(G/S) <= d(G)

The first line of the proof read like,

By the correspondence theorem, p(G/S) = (pG +S)/S,

How is this equation derived? As the correspondence theorem mainly states isomorphism, I cannot see where there is equation involved? It would be greatly appreciated if anyone could help on this. Many thanks!

2. Jun 17, 2012

### DonAntonio

In my book (3rd edition), it is chapter 6, lemma 6.10 (i).

Now, $\,p(G/S)\,:=\{p(x+S)=px +S\;|\;x\in G\}\leq G/S$ , and since this is a subgroup of the quotient $\,G/S\,$, the

correspondence theorem tells us that there exists $\,H\leq G\,\,s.t.\,\,p(G/S)=H/S\,$ , and it's not hard to realize that in fact

$\,H=pG+S\,$ , for example $\,\forall x\in G\,\,,\, px + S\in p\left(G/S\right)\Longrightarrow pG+s\leq p(G/S)\,$. Now you try to prove the other way around.

DonAntonio

Ps The correspondence theorem is *not* about isomorphisms merely but about a 1-1 correspondence between

subgroups of $\,G/N\,$ and subgroups of $G\,$ containing $\,N\,$.

3. Jun 17, 2012

### jz2012

Thanks a lot DonAntonio,

$\,p(G/S)\,:=\{p(x+S)=px +S\;|\;x\in G\}$ is this a definition for p(G/S), not
$\,p(G/S)\,:=\{p(x+S)=px +pS = px+ S\;|\;x\in G\}$ (i assume := means definition?)