MHB Correspondence Theorem for Groups - Yet Another Question

[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 have another question regarding the proof of Proposition 2.123 part (i) ... which I think is necessary in order to understand the corresponding Proposition for rings ... (apologies to Euge if he answered this question in a previous post ... but I am still reflecting on Euge's post ... ... )Proposition 2.123 part (i) and its proof reads as follows:View attachment 4721In the proof of the Proposition above, we read the following:

" (i) Let $$\Phi \ : \ Sub(G; K) \rightarrow Sub(G/K)$$ denote the function $$\Phi \ : \ S \mapsto S/K$$ (it is routine to check that if $$S$$ is a subgroup of $$G$$ containing $$K$$, then $$S/K$$ is a subgroup of $$G/K$$) ... ... "

Can someone please explain why in the above context, that if $$S$$ is a subgroup of $$G$$ containing $$K$$, then $$S/K$$ is a subgroup of $$G/K$$ ... ...

Hope someone can clarify why this is the case ...

Peter

 

[Euge]

Hi Peter,

Suppose $S$ is a subgroup of $G$ which contains $K$. Since $K$ is normal in $G$, $gkg^{-1} \in K$ for all $g\in G$. Since every $s\in S$ belongs to $G$ (as $S$ is a subset of $G$), in particular $sks^{-1}\in K$ for all $s\in S$. Hence, $K$ is normal in $S$ and we can form the factor group $S/K$. Now since $S$ is a subset of $G$, then $S/K$ is a subset of $G/K$. So since $S/K$ is a subset of $G/K$ that is a group in its own right, $S/K$ is a subgroup of $G/K$.

 

[Math Amateur]

Hi Peter,

Suppose $S$ is a subgroup of $G$ which contains $K$. Since $K$ is normal in $G$, $gkg^{-1} \in K$ for all $g\in G$. Since every $s\in S$ belongs to $G$ (as $S$ is a subset of $G$), in particular $sks^{-1}\in K$ for all $s\in S$. Hence, $K$ is normal in $S$ and we can form the factor group $S/K$. Now since $S$ is a subset of $G$, then $S/K$ is a subset of $G/K$. So since $S/K$ is a subset of $G/K$ that is a group in its own right, $S/K$ is a subgroup of $G/K$.

Thanks Euge ... very clear and easy to follow ... a real help ... thank you ...

Peter

 

[Deveno]

One can show that $S/K$ forms a group in its own right:

$S/K = \{sK: s \in S\}$.

Closure: let $s_1K,s_2K \in S/K$. Then $(s_1K)(s_2K) = (s_1s_2)K$ (by definition of coset multiplication, which we can do for any two elements of the SET $S/K \subseteq G/K$). Since $s_1s_2 \in S$ (since $S$ is a subgroup of $G$), we have: $(s_1s_2)K \in S/K$.

Existence of inverses: for any $sK \in S/K$, we have in $G/K$ that $(sK)^{-1} = s^{-1}K$. Since $S$ is a subgroup of $G$, we have: $s^{-1} \in S$, and thus $s^{-1}K \in S/K$.

The proviso that $K \subseteq S$ is so that the set $S/K$ as defined above (which we could form regardless as the image under the canonical homomorphism $G \to G/K$) has the same meaning as the group $S/K$ as an independent entity, which only makes sense if $K$ is a (normal) subgroup of $S$.

 

[Math Amateur]

One can show that $S/K$ forms a group in its own right:

$S/K = \{sK: s \in S\}$.

Closure: let $s_1K,s_2K \in S/K$. Then $(s_1K)(s_2K) = (s_1s_2)K$ (by definition of coset multiplication, which we can do for any two elements of the SET $S/K \subseteq G/K$). Since $s_1s_2 \in S$ (since $S$ is a subgroup of $G$), we have: $(s_1s_2)K \in S/K$.

Existence of inverses: for any $sK \in S/K$, we have in $G/K$ that $(sK)^{-1} = s^{-1}K$. Since $S$ is a subgroup of $G$, we have: $s^{-1} \in S$, and thus $s^{-1}K \in S/K$.

The proviso that $K \subseteq S$ is so that the set $S/K$ as defined above (which we could form regardless as the image under the canonical homomorphism $G \to G/K$) has the same meaning as the group $S/K$ as an independent entity, which only makes sense if $K$ is a (normal) subgroup of $S$.

Thanks Deveno ... a really helpful post ...

Peter

- - - Updated - - -

One can show that $S/K$ forms a group in its own right:

$S/K = \{sK: s \in S\}$.

Closure: let $s_1K,s_2K \in S/K$. Then $(s_1K)(s_2K) = (s_1s_2)K$ (by definition of coset multiplication, which we can do for any two elements of the SET $S/K \subseteq G/K$). Since $s_1s_2 \in S$ (since $S$ is a subgroup of $G$), we have: $(s_1s_2)K \in S/K$.

Existence of inverses: for any $sK \in S/K$, we have in $G/K$ that $(sK)^{-1} = s^{-1}K$. Since $S$ is a subgroup of $G$, we have: $s^{-1} \in S$, and thus $s^{-1}K \in S/K$.

The proviso that $K \subseteq S$ is so that the set $S/K$ as defined above (which we could form regardless as the image under the canonical homomorphism $G \to G/K$) has the same meaning as the group $S/K$ as an independent entity, which only makes sense if $K$ is a (normal) subgroup of $S$.

Thanks Deveno ... a really helpful post ...

Peter

 

[Math Amateur]

One can show that $S/K$ forms a group in its own right:

$S/K = \{sK: s \in S\}$.

Closure: let $s_1K,s_2K \in S/K$. Then $(s_1K)(s_2K) = (s_1s_2)K$ (by definition of coset multiplication, which we can do for any two elements of the SET $S/K \subseteq G/K$). Since $s_1s_2 \in S$ (since $S$ is a subgroup of $G$), we have: $(s_1s_2)K \in S/K$.

Existence of inverses: for any $sK \in S/K$, we have in $G/K$ that $(sK)^{-1} = s^{-1}K$. Since $S$ is a subgroup of $G$, we have: $s^{-1} \in S$, and thus $s^{-1}K \in S/K$.

The proviso that $K \subseteq S$ is so that the set $S/K$ as defined above (which we could form regardless as the image under the canonical homomorphism $G \to G/K$) has the same meaning as the group $S/K$ as an independent entity, which only makes sense if $K$ is a (normal) subgroup of $S$.

Hi Deveno,

I have been reflecting on your post ... and now need some further help ...

How exactly do we know that in $G/K$ that $(sK)^{-1} = s^{-1}K$?

Peter

 

[Deveno]

Inverses in a group are unique, and:

$(sK)(s^{-1}K) = (ss^{-1})K = eK = K = e_{G/K}$, so

$(sK)^{-1}$ must be $s^{-1}K$.

 

[Math Amateur]

Inverses in a group are unique, and:

$(sK)(s^{-1}K) = (ss^{-1})K = eK = K = e_{G/K}$, so

$(sK)^{-1}$ must be $s^{-1}K$.

Oh yes, indeed ... should have seen that ...

Thanks for your help, Deveno ...

Peter