Correspondence Theorem for Groups - Rotman, Proposition

[Math Amateur]

I am reading Joseph J. Rotman's book: A First Course in Abstract Algebra with Applications (Third Edition) ...

I am currently revising Section 2.6 Quotient Groups in order to understand rings better ...

I need help with understanding the proof of Proposition 2.123 part (i) ... which I think is necessary in order to understand the corresponding Proposition for rings ...

Proposition 2.123 part (i) and its proof reads as follows:View attachment 4716I am having real difficulties understanding the proof of part (i) and indeed seeing the big picture or strategy of the proof ... (if someone can provide a clearer proof then I would be really grateful ...)

I am confused over the proof and am finding it difficult to frame sensible questions about the above proof ... but I'll try to explain my confusions ...

\(\displaystyle \Phi\) is a mapping from a set \(\displaystyle S_i\) (containing \(\displaystyle K\)) to a set \(\displaystyle S_i / K\) ... that is from a set \(\displaystyle S_i\) containing \(\displaystyle K\) to the set of cosets of \(\displaystyle S_i \text{ mod } K\) ...

\(\displaystyle \pi\) is the natural map \(\displaystyle \pi \ : \ s_i \rightarrow S_i / K\) where \(\displaystyle \pi(a) = a + K = [a]\) , the coset containing the element \(\displaystyle a\) ...

BUT ... what is \(\displaystyle \pi(S)\) and why, exactly does \(\displaystyle \pi^{-1} \pi (S)\) imply the injectivity of \(\displaystyle \Phi\) ... ...
A second question I have is as follows:

The above text includes the following statement:

" ... ... let \(\displaystyle a \in \pi^{-1} \pi(S)\), so that \(\displaystyle \pi(a) = \pi(s)\) for some \(\displaystyle s \in S\). It follows that \(\displaystyle as^{-1} \in \text{ ker } \pi = K\), so that \(\displaystyle a = sk\) for some \(\displaystyle k \in K\) ... ... "

My question is as follows:

How does \(\displaystyle a \in \pi^{-1} \pi(S)\), (so that \(\displaystyle \pi(a) = \pi(s)\) for some \(\displaystyle s \in S\)) ... ... imply that \(\displaystyle as^{-1} \in \text{ ker } \pi = K\) ... ... and, further how does \(\displaystyle as^{-1} \in \text{ ker } \pi = K\) ... ... imply that \(\displaystyle a = sk\) for some \(\displaystyle k \in K\) ...

Hope someone can help ... ...

Peter*** NOTE *** ... ... if someone can provide a clear proof of this Proposition ... plus perhaps a relevant example ... ... I would be extremely grateful ...

*** EDIT *** Have been reflecting on my issues and checking in Rotman's text for answers ...

I now believe that \(\displaystyle \pi(S)\) is the direct image of S under \pi ...

Further Proposition 2.14 shows that in the case where \(\displaystyle K \leq S \leq G\) then \(\displaystyle \pi\) is an injection, then \(\displaystyle \pi^{-1} \pi = S\) ... BUT ... in the above Proposition, Rotman seems to be assuming that the implication also works in the other direction ... that is that \(\displaystyle \pi^{-1} \pi = S\) implies that \(\displaystyle \pi\) is injective ... ... so I am still confused ... ...Proposition 2.14 reads as follows:

View attachment 4719By the way it appears to me that there is a typo in part (iii) ... I think it should read as follows:

" ... ... ...

(iii) If \(\displaystyle T \subseteq X\), then \(\displaystyle T \subseteq f^{-1}f(T)\); if \(\displaystyle f\) is an injection and \(\displaystyle T \subseteq X\), then \(\displaystyle T = f^{-1}f(T)\)."

But then my problem ... as stated above is that Rotman appears to be using, not the stated result of part (iii) of Proposition 2.14 ... but instead uses the following result ... ...

If \(\displaystyle T = f^{-1}f(T)\) and \(\displaystyle T \subseteq X\) then \(\displaystyle f\) is injective ... ...

Hope someone can clarify this matter ...

 

[Euge]

Let's begin with an example. Consider the sequence of inclusions $10 \Bbb Z \hookrightarrow 5\Bbb Z\hookrightarrow \Bbb Z$. Factoring each of these groups out by $10 \Bbb Z$, we get $0 \hookrightarrow 5\Bbb Z_{10} \hookrightarrow \Bbb Z_{10}$. To revert back to the original sequence, we consider "pulling back" the subgroups $0$, $5\Bbb Z_{10}$, and $\Bbb Z_{10}$. Let $\pi : \Bbb Z \to \Bbb Z_{10}$, $\pi(x) = [x]_{10}$ be the natural projection. If $H$ is a subgroup of $\Bbb Z_{10}$, then $\pi^{-1}(H)$ is a subgroup of $\Bbb Z$; we have pulled back a subgroup of $\Bbb Z_{10}$ to a subgroup of $\Bbb Z$. We know that $\pi^{-1}(\Bbb Z_{10}) = \Bbb Z$, but how about $\pi^{-1}(5\Bbb Z_{10})$ and $\pi^{-1}(0)$? Well, $\pi^{-1}(0)$ consist of all $x\in \Bbb Z$ such that $[x]_{10} = [0]_{10}$, i.e., $x$ is a multiple of $10$. Therefore, $\pi^{-1}(0) = 10\Bbb Z$. Now $\pi^{-1}(5\Bbb Z_{10})$ consists of those elements $x\in \Bbb Z$ for which $[x]_{10} \in 5\Bbb Z_{10}$. Now $5\Bbb Z_{10}$ is generated by $[5]_{10}$, and $\pi$ maps $5$ to that generator, so $5\Bbb Z = \pi^{-1}(5\Bbb Z_{10})$. So indeed, the sequence $\pi^{-1}(0) \hookrightarrow \pi^{-1}(5\Bbb Z_{10}) \hookrightarrow \pi^{-1}(\Bbb Z_{10})$ is exactly the sequence $10 \Bbb Z \hookrightarrow 5\Bbb Z \hookrightarrow 10 \Bbb Z$ that we started with!

Part (i) of the Correspondence Theorem (or Lattice Isomorphism Theorem, as some may call it) may be proven by showing that $\Phi$ is well-defined and admits an inverse. Consider the function $\Psi : \mathrm{Sub}(G/K) \to \mathrm{Sub}(G;K)$, $\Psi(N) = \pi^{-1}(N)$. For any $N\in \mathrm{Sub}(G/K)$, $\Phi(\Psi(N)) = \pi^{-1}(N)/K$, which I claim to be $N$. Indeed, $$[x]\in \pi^{-1}(N)/K \iff x\in \pi^{-1}(N) \iff \pi(x)\in N \iff [x]\in N.$$ Now $\Psi(\Phi(S)) = S$ for all $S\in \mathrm{Sub}(G;K)$. For given $S\in \mathrm{Sub}(G;K)$, $\Psi(\Phi(S)) = \pi^{-1}(S/K)$, and $$x\in \pi^{-1}(S/K) \iff \pi(x) \in S/K \iff [x]\in S/K \iff x\in S.$$ Hence $\Psi(\Phi(S)) = S$, and consequently $\Psi$ is the inverse of $\Phi$.

 

[Math Amateur]

Let's begin with an example. Consider the sequence of inclusions $10 \Bbb Z \hookrightarrow 5\Bbb Z\hookrightarrow \Bbb Z$. Factoring each of these groups out by $10 \Bbb Z$, we get $0 \hookrightarrow 5\Bbb Z_{10} \hookrightarrow \Bbb Z_{10}$. To revert back to the original sequence, we consider "pulling back" the subgroups $0$, $5\Bbb Z_{10}$, and $\Bbb Z_{10}$. Let $\pi : \Bbb Z \to \Bbb Z_{10}$, $\pi(x) = [x]_{10}$ be the natural projection. If $H$ is a subgroup of $\Bbb Z_{10}$, then $\pi^{-1}(H)$ is a subgroup of $\Bbb Z$; we have pulled back a subgroup of $\Bbb Z_{10}$ to a subgroup of $\Bbb Z$. We know that $\pi^{-1}(\Bbb Z_{10}) = \Bbb Z$, but about $\pi^{-1}(5\Bbb Z_{10})$ and $\pi^{-1}(0)$? Well, $\pi^{-1}(0)$ consist of all $x\in \Bbb Z$ such that $[x]_{10} = [0]_{10}$, i.e., $x$ is a multiple of $10$. Therefore, $\pi^{-1}(0) = 10\Bbb Z$. Now $\pi^{-1}(5\Bbb Z_{10})$ consists of those elements $x\in \Bbb Z$ for which $[x]_{10} \in 5\Bbb Z_{10}$. Now $5\Bbb Z_{10}$ is generated by $[5]_{10}$, and $\pi$ maps $5$ to that generator, so $5\Bbb Z = \pi^{-1}(5\Bbb Z_{10})$. So indeed, the sequence $\pi^{-1}(0) \hookrightarrow \pi^{-1}(5\Bbb Z_{10}) \hookrightarrow \pi^{-1}(\Bbb Z_{10})$ is exactly the sequence $10 \Bbb Z \hookrightarrow 5\Bbb Z \hookrightarrow 10 \Bbb Z$ that we started with!

Part (i) of the Correspondence Theorem (or Lattice Isomorphism Theorem, as some may call it) may be proven by showing that $\Phi$ is well-defined and admits an inverse. Consider the function $\Psi : \mathrm{Sub}(G/K) \to \mathrm{Sub}(G;K)$, $\Psi(N) = \pi^{-1}(N)$. For any $N\in \mathrm{Sub}(G/K)$, $\Phi(\Psi(N)) = \pi^{-1}(N)/K$, which I claim to be $N$. Indeed, $$[x]\in \pi^{-1}(N)/K \iff x\in \pi^{-1}(N) \iff \pi(x)\in N \iff [x]\in N.$$ Now $\Psi(\Phi(S)) = S$ for all $S\in \mathrm{Sub}(G;K)$. For given $S\in \mathrm{Sub}(G;K)$, $\Psi(\Phi(S)) = \pi^{-1}(S/K)$, and $$x\in \pi^{-1}(S/K) \iff \pi(x) \in S/K \iff [x]\in S/K \iff x\in S.$$ Hence $\Psi(\Phi(S)) = S$, and consequently $\Psi$ is the inverse of $\Phi$.

Thanks for the help, Euge ... just working through your post ... BUT ... just a quick clarification ...

You write:

"Let's begin with an example. Consider the sequence of inclusions $10 \Bbb Z \hookrightarrow 5\Bbb Z\hookrightarrow \Bbb Z$. ... ... "

What is the meaning of the symbol "\(\displaystyle \hookrightarrow\)" ?

Is it the same as set inclusion ... that is \(\displaystyle \subseteq\) ?

Peter

 

[Euge]

Yes, here I mean set inclusion.

 

[Math Amateur]

Yes, here I mean set inclusion.

Hi Euge,

Just working through your post ...

Thanks so much for your help ...

Peter

 

[Math Amateur]

Let's begin with an example. Consider the sequence of inclusions $10 \Bbb Z \hookrightarrow 5\Bbb Z\hookrightarrow \Bbb Z$. Factoring each of these groups out by $10 \Bbb Z$, we get $0 \hookrightarrow 5\Bbb Z_{10} \hookrightarrow \Bbb Z_{10}$. To revert back to the original sequence, we consider "pulling back" the subgroups $0$, $5\Bbb Z_{10}$, and $\Bbb Z_{10}$. Let $\pi : \Bbb Z \to \Bbb Z_{10}$, $\pi(x) = [x]_{10}$ be the natural projection. If $H$ is a subgroup of $\Bbb Z_{10}$, then $\pi^{-1}(H)$ is a subgroup of $\Bbb Z$; we have pulled back a subgroup of $\Bbb Z_{10}$ to a subgroup of $\Bbb Z$. We know that $\pi^{-1}(\Bbb Z_{10}) = \Bbb Z$, but how about $\pi^{-1}(5\Bbb Z_{10})$ and $\pi^{-1}(0)$? Well, $\pi^{-1}(0)$ consist of all $x\in \Bbb Z$ such that $[x]_{10} = [0]_{10}$, i.e., $x$ is a multiple of $10$. Therefore, $\pi^{-1}(0) = 10\Bbb Z$. Now $\pi^{-1}(5\Bbb Z_{10})$ consists of those elements $x\in \Bbb Z$ for which $[x]_{10} \in 5\Bbb Z_{10}$. Now $5\Bbb Z_{10}$ is generated by $[5]_{10}$, and $\pi$ maps $5$ to that generator, so $5\Bbb Z = \pi^{-1}(5\Bbb Z_{10})$. So indeed, the sequence $\pi^{-1}(0) \hookrightarrow \pi^{-1}(5\Bbb Z_{10}) \hookrightarrow \pi^{-1}(\Bbb Z_{10})$ is exactly the sequence $10 \Bbb Z \hookrightarrow 5\Bbb Z \hookrightarrow 10 \Bbb Z$ that we started with!

Part (i) of the Correspondence Theorem (or Lattice Isomorphism Theorem, as some may call it) may be proven by showing that $\Phi$ is well-defined and admits an inverse. Consider the function $\Psi : \mathrm{Sub}(G/K) \to \mathrm{Sub}(G;K)$, $\Psi(N) = \pi^{-1}(N)$. For any $N\in \mathrm{Sub}(G/K)$, $\Phi(\Psi(N)) = \pi^{-1}(N)/K$, which I claim to be $N$. Indeed, $$[x]\in \pi^{-1}(N)/K \iff x\in \pi^{-1}(N) \iff \pi(x)\in N \iff [x]\in N.$$ Now $\Psi(\Phi(S)) = S$ for all $S\in \mathrm{Sub}(G;K)$. For given $S\in \mathrm{Sub}(G;K)$, $\Psi(\Phi(S)) = \pi^{-1}(S/K)$, and $$x\in \pi^{-1}(S/K) \iff \pi(x) \in S/K \iff [x]\in S/K \iff x\in S.$$ Hence $\Psi(\Phi(S)) = S$, and consequently $\Psi$ is the inverse of $\Phi$.

Hi Euge,

Thanks again for the help ... but ... I need a bit more help/clarification ...

You write:

" ... ... Part (i) of the Correspondence Theorem (or Lattice Isomorphism Theorem, as some may call it) may be proven by showing that $\Phi$ is well-defined and admits an inverse. ... ... "I know that if \(\displaystyle \Phi\) is a group homomorphism then if it admits an inverse it is an isomorphism ... and hence, then, a bijection ... so I can see where you are going with proving \(\displaystyle \Phi\) is a bijection ... but I am a bit confused on the following point ...

How do we know that \(\displaystyle \Phi\) is a group homomorphism ... surely we need to show that ...

... for groups \(\displaystyle S,T \in Sub(G; K)\) we have \(\displaystyle \Phi(ST) = \Phi(S) \Phi(T)\) ... ...

... that is we need to show \(\displaystyle ST/K = (S/K)(T/K)\) ...

But what does the above expression mean ... that is what does \(\displaystyle ST\) mean when \(\displaystyle S\) and \(\displaystyle T\) are groups ... and, if it is meaningful, how do we prove it true ...?

Hope you can help ...

Peter

 

[Euge]

Peter, the goal was to show that $\Phi$ is a one-to-one correspondence, not an isomorphism - $\mathrm{Sub}(G;K)$ and $\mathrm{Sub}(G/K)$ are in general sets, not groups.

 

[Math Amateur]

Peter, the goal was to show that $\Phi$ is a one-to-one correspondence, not an isomorphism - $\mathrm{Sub}(G;K)$ and $\mathrm{Sub}(G/K)$ are in general sets, not groups.

Oh my goodness ... indeed you are right, of course ...

Sorry for the momentary confusion ...

Peter

 

[Math Amateur]

Let's begin with an example. Consider the sequence of inclusions $10 \Bbb Z \hookrightarrow 5\Bbb Z\hookrightarrow \Bbb Z$. Factoring each of these groups out by $10 \Bbb Z$, we get $0 \hookrightarrow 5\Bbb Z_{10} \hookrightarrow \Bbb Z_{10}$. To revert back to the original sequence, we consider "pulling back" the subgroups $0$, $5\Bbb Z_{10}$, and $\Bbb Z_{10}$. Let $\pi : \Bbb Z \to \Bbb Z_{10}$, $\pi(x) = [x]_{10}$ be the natural projection. If $H$ is a subgroup of $\Bbb Z_{10}$, then $\pi^{-1}(H)$ is a subgroup of $\Bbb Z$; we have pulled back a subgroup of $\Bbb Z_{10}$ to a subgroup of $\Bbb Z$. We know that $\pi^{-1}(\Bbb Z_{10}) = \Bbb Z$, but how about $\pi^{-1}(5\Bbb Z_{10})$ and $\pi^{-1}(0)$? Well, $\pi^{-1}(0)$ consist of all $x\in \Bbb Z$ such that $[x]_{10} = [0]_{10}$, i.e., $x$ is a multiple of $10$. Therefore, $\pi^{-1}(0) = 10\Bbb Z$. Now $\pi^{-1}(5\Bbb Z_{10})$ consists of those elements $x\in \Bbb Z$ for which $[x]_{10} \in 5\Bbb Z_{10}$. Now $5\Bbb Z_{10}$ is generated by $[5]_{10}$, and $\pi$ maps $5$ to that generator, so $5\Bbb Z = \pi^{-1}(5\Bbb Z_{10})$. So indeed, the sequence $\pi^{-1}(0) \hookrightarrow \pi^{-1}(5\Bbb Z_{10}) \hookrightarrow \pi^{-1}(\Bbb Z_{10})$ is exactly the sequence $10 \Bbb Z \hookrightarrow 5\Bbb Z \hookrightarrow 10 \Bbb Z$ that we started with!

Part (i) of the Correspondence Theorem (or Lattice Isomorphism Theorem, as some may call it) may be proven by showing that $\Phi$ is well-defined and admits an inverse. Consider the function $\Psi : \mathrm{Sub}(G/K) \to \mathrm{Sub}(G;K)$, $\Psi(N) = \pi^{-1}(N)$. For any $N\in \mathrm{Sub}(G/K)$, $\Phi(\Psi(N)) = \pi^{-1}(N)/K$, which I claim to be $N$. Indeed, $$[x]\in \pi^{-1}(N)/K \iff x\in \pi^{-1}(N) \iff \pi(x)\in N \iff [x]\in N.$$ Now $\Psi(\Phi(S)) = S$ for all $S\in \mathrm{Sub}(G;K)$. For given $S\in \mathrm{Sub}(G;K)$, $\Psi(\Phi(S)) = \pi^{-1}(S/K)$, and $$x\in \pi^{-1}(S/K) \iff \pi(x) \in S/K \iff [x]\in S/K \iff x\in S.$$ Hence $\Psi(\Phi(S)) = S$, and consequently $\Psi$ is the inverse of $\Phi$.

Thanks so much Euge ...

Have now followed your proof ...

... your proof of is so much clearer than Rotman's proof which to me remains a mystery!

Thanks again,

Peter