# The Correspondence Theorem for Groups .... Rotman, Proposition 1.82 ....

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In summary, the author is discussing the correspondence theorem in Proposition 1.82. Peter needs help understanding the proof and asks the author for clarification. The author explains how the theorem works based on the concept of an identity in a group and the equivalence of as^{-1} and k.
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I am reading the book: "Advanced Modern Algebra" (Second Edition) by Joseph J. Rotman ...

I am currently focused on Chapter 1: Groups I ...

I need help with an aspect of the proof of Proposition 1.82 (Correspondence Theorem) ...

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View attachment 7994
In the above proof by Rotman we read the following:

" ... ... For the reverse inclusion let $$\displaystyle a \in \pi^{-1} \pi (S)$$, so that $$\displaystyle \pi (a) = \pi (s)$$ for some $$\displaystyle s \in S$$. It follows that $$\displaystyle as^{-1} \in \text{ ker } \pi = K$$ ... ... "Can someone please explain exactly how/why $$\displaystyle \pi (a) = \pi (s)$$ implies that $$\displaystyle as^{-1} \in \text{ ker } \pi = K$$ ... ... Peter
===========================================================================================***EDIT***

Just had some thoughts ... BUT ... unfortunately my logic does not seem to line up with Rotman's logic ...

My thoughts are as follows:

$$\displaystyle \pi (a) = \pi (s)$$

$$\displaystyle \Longrightarrow aK = sK$$

$$\displaystyle \Longrightarrow a = sk$$ for some $$\displaystyle k \in K$$ since $$\displaystyle a$$ must belong to $$\displaystyle sK$$ ... ... (is this a legitimate step ...)

$$\displaystyle \Longrightarrow s^{-1} a = k$$

$$\displaystyle \Longrightarrow as^{-1} = k$$

$$\displaystyle \Longrightarrow as^{-1} \in \text{ker } \pi = K$$ Is that correct?

BUT note ... logic is different from Rotman's set of steps ... ... what is Rotman's logic ...?Peter

Last edited:
Hi, Peter.

The reasoning in your EDIT section copied below is correct, including the step where you ask if $a=sk$ is legitimate (it is because $K$ is a (sub)group and so must contain the identity, which implies $a\in aK$).

Peter said:
Just had some thoughts ... BUT ... unfortunately my logic does not seem to line up with Rotman's logic ...

My thoughts are as follows:

$$\displaystyle \pi (a) = \pi (s)$$

$$\displaystyle \Longrightarrow aK = sK$$

$$\displaystyle \Longrightarrow a = sk$$ for some $$\displaystyle k \in K$$ since $$\displaystyle a$$ must belong to $$\displaystyle sK$$ ... ... (is this a legitimate step ...)

$$\displaystyle \Longrightarrow s^{-1} a = k$$

$$\displaystyle \Longrightarrow as^{-1} = k$$

$$\displaystyle \Longrightarrow as^{-1} \in \text{ker } \pi = K$$ Is that correct?

BUT note ... logic is different from Rotman's set of steps ... ... what is Rotman's logic ...?

Indeed, your reasoning is expressed a little different from Rotman's (with yours on the level of elements from $G$ throughout and with Rotman's given on the level of cosets and group homomorphisms), but it is equally valid and yields the same conclusion, well done!

Peter said:
Can someone please explain exactly how/why $$\displaystyle \pi (a) = \pi (s)$$ implies that $$\displaystyle as^{-1} \in \text{ ker } \pi = K$$ ... ...

For any group homomorphism, say $h:G_{1}\rightarrow G_{2},$ $h(g^{-1})=\left[h(g)\right]^{-1}$ (a short but good exercise to try for yourself if you're not already familiar with this fact). Hence, $\pi(a)=\pi(s)\Longrightarrow\left[\pi(a)\right]^{-1}=\left[\pi(s)\right]^{-1}.$

Now we compute: $\pi(as^{-1})=\pi(a)\pi(s^{-1})=\pi(a)\left[\pi(s)\right]^{-1}=\pi(a)\left[\pi(a)\right]^{-1}=\text{Identity in}~G/K=K.$

GJA said:
Hi, Peter.

The reasoning in your EDIT section copied below is correct, including the step where you ask if $a=sk$ is legitimate (it is because $K$ is a (sub)group and so must contain the identity, which implies $a\in aK$).
Indeed, your reasoning is expressed a little different from Rotman's (with yours on the level of elements from $G$ throughout and with Rotman's given on the level of cosets and group homomorphisms), but it is equally valid and yields the same conclusion, well done!
For any group homomorphism, say $h:G_{1}\rightarrow G_{2},$ $h(g^{-1})=\left[h(g)\right]^{-1}$ (a short but good exercise to try for yourself if you're not already familiar with this fact). Hence, $\pi(a)=\pi(s)\Longrightarrow\left[\pi(a)\right]^{-1}=\left[\pi(s)\right]^{-1}.$

Now we compute: $\pi(as^{-1})=\pi(a)\pi(s^{-1})=\pi(a)\left[\pi(s)\right]^{-1}=\pi(a)\left[\pi(a)\right]^{-1}=\text{Identity in}~G/K=K.$
Thanks GJA ...

Peter

## What is the Correspondence Theorem for Groups?

The Correspondence Theorem for Groups, also known as the Lattice Isomorphism Theorem, states that for any normal subgroup N of a group G, there exists a bijection between the subgroups of G that contain N and the subgroups of G/N. This theorem is a fundamental result in group theory and has many applications in algebra and geometry.

## How is the Correspondence Theorem for Groups proven?

The Correspondence Theorem for Groups is proven using the concept of cosets and quotient groups. First, it is shown that the set of subgroups of G that contain N is in bijection with the set of cosets of N in G. Then, it is shown that the set of subgroups of G/N is in bijection with the set of subgroups of G that contain N. Combining these two bijections gives the desired result.

## What is the significance of Proposition 1.82 in Rotman's book?

Proposition 1.82 in Rotman's book is a special case of the Correspondence Theorem for Groups, where N is taken to be the trivial subgroup {e}. This proposition states that for any subgroup H of a group G, there exists a bijection between the subgroups of G that contain H and the subgroups of G/H. This result is important because it allows us to study the structure of a group by considering its quotient groups.

## How is the Correspondence Theorem for Groups used in group theory?

The Correspondence Theorem for Groups has many important applications in group theory. It allows us to classify all the subgroups of a group, determine the normal subgroups of a group, and understand the structure of quotient groups. It also plays a key role in the study of group homomorphisms and factor groups.

## Can the Correspondence Theorem for Groups be extended to other algebraic structures?

Yes, the Correspondence Theorem for Groups can be extended to other algebraic structures such as rings and modules. In fact, the concept of quotient structures and the corresponding isomorphism theorems are important in many areas of algebra and beyond. The Correspondence Theorem for Groups is just one example of this powerful tool in mathematics.

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