P primary group and the correspondence theorem

[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!

 

[DonAntonio]

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!

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 $\,N\,$.

 

[jz2012]

Thanks a lot DonAntonio,

this is very helpful!

I am just wondering
$\,p(G/S)\,:=\{p(x+S)=px +S\;|\;x\in G\}$ 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?)