Correspondence Theorem for Modules - Rotman, Section 6.1

[Math Amateur]

I am reading Joseph J. Rotman's book: Advanced Modern Algebra and I am currently focused on Section 6.1 Modules ...

I need some help with the proof of Theorem 6.22 (Correspondence Theorem) ... ...

Theorem 6.22 and its proof read as follows:http://mathhelpboards.com/attachment.php?attachmentid=4921&stc=1In the above proof we read:

" ... ... Since every module is an additive abelian group, every submodule is a subgroup, and so the Correspondence Theorem for Groups, Theorem 1.82, shows that \(\displaystyle \phi\) is an injection that preserves inclusions: \(\displaystyle S \subseteq S'\) in \(\displaystyle M\) if and only if \(\displaystyle S/T \subseteq S'/T\) in \(\displaystyle M/T\). ... ... "How can we deduce this straight from a Theorem on groups ... ... ? ... how do we know it automatically holds for the action of the ring \(\displaystyle R\) on \(\displaystyle M\) and the laws the action or scalar multiplication must follow ...

Further ... can someone indicate how the rest of the proof would read ...

Hope someone can help ...

Peter

=======================================================*** EDIT 1 ***

The above text from Rotman's Advanced Modern Algebra (AMA) references the Correspondence Theorem for Groups ... so to permit MHB readers to understand the full context of this post, I am providing the Correspondence Theorem for Groups as it reads in AMA ... as follows:View attachment 4922
View attachment 4923

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*** EDIT 2 ***

The above text from Rotman's Advanced Modern Algebra (AMA) also references the Correspondence Theorem for Rings (Proposition 5.1) ... so to permit MHB readers to understand the full context of this post, I am providing the Correspondence Theorem for Rings as it reads in AMA ... as follows:View attachment 4924

 

[Deveno]

Well, we already know the "group part". Namely, if $f:M \to N$ is a surjective $R$-module homomorphism, then it is (among other things) a group homomorphism (of the additive groups $(M,+)$ onto $(N,+)$).

But an $R$-module homomorphism (by definition) respects the scalar multiplication, so we have, for any submodule $M'$ of $M$, that $f(M')$ is a submodule of $N$.

Why?

Well, we know a priori that $f(M')$ is an abelian group under addition. Thus it suffices to show that for any $n \in f(M')$ and any $a \in R$ that $an \in f(M')$.

Since $M'$ is an $R$-submodule of $M$, we know that for any $m \in M'$, and any $a \in R$, that $am \in M'$. Now $n \in f(M')$, so $n = f(m')$ for some $m' \in M'$. Thus for any $a \in R$, we have $am' \in M'$.

So...$an = af(m') = f(am')$ (since $f$ is an $R$-module homomorphism)

and since $am' \in M'$, this shows $an \in f(M')$, so $f(M')$ is indeed an $R$-submodule of $N$.