Second Isomorphism Theorem for Groups

[Math Amateur]

I am revising the Isomorphism Theorems for Groups in order to better understand the Isomorphism Theorems for Modules.

I need some help in understanding Dummit and Foote's proof of the Second Isomorphism Theorems for Groups (Diamond Isomorphism Theorem ? why Diamond ?).

The relevant text from D&F is as follows:
View attachment 3270
In the proof above we read:

"Proof: By Corollary 15, \(\displaystyle AB\) is a subgroup of \(\displaystyle G\).

Since \(\displaystyle A \leq N_G (B)\) by assumption and \(\displaystyle B \leq N_G (B)\) trivially ...

... it follows that \(\displaystyle AB \leq N_G (B)\) i.e. \(\displaystyle B\) is a normal subgroup of the subgroup \(\displaystyle AB\) ... ... ... "

Can someone please explain to me why \(\displaystyle AB \leq N_G (B)\) means that \(\displaystyle B\) is a normal subgroup of the subgroup \(\displaystyle AB\)?

Peter

 

[Math Amateur]

I am revising the Isomorphism Theorems for Groups in order to better understand the Isomorphism Theorems for Modules.

I need some help in understanding Dummit and Foote's proof of the Second Isomorphism Theorems for Groups (Diamond Isomorphism Theorem ? why Diamond ?).

The relevant text from D&F is as follows:
View attachment 3270
In the proof above we read:

"Proof: By Corollary 15, \(\displaystyle AB\) is a subgroup of \(\displaystyle G\).

Since \(\displaystyle A \leq N_G (B)\) by assumption and \(\displaystyle B \leq N_G (B)\) trivially ...

... it follows that \(\displaystyle AB \leq N_G (B)\) i.e. \(\displaystyle B\) is a normal subgroup of the subgroup \(\displaystyle AB\) ... ... ... "

Can someone please explain to me why \(\displaystyle AB \leq N_G (B)\) means that \(\displaystyle B\) is a normal subgroup of the subgroup \(\displaystyle AB\)?

Peter

I now have developed an answer to my own question ... ... BUT now have a second worry with D&F's proof ...

First I present my answer ...

Following some basic reading and reflection it now appears to me that the answer to my question:

"why \(\displaystyle AB \leq N_G (B)\) means that \(\displaystyle B\) is a normal subgroup of the subgroup \(\displaystyle AB\)?"

follows basically from the definitions of normal subgroup and normalizer ...

... but I would be grateful if someone would confirm that my argument/analysis below is correct ...
Now ... ... following D&F's definition of a normal group on page 82 we have that:

A subgroup \(\displaystyle B\) of a group \(\displaystyle AB\) is called normal if every element of \(\displaystyle AB\) normalizes the subgroup \(\displaystyle B\) ... ...

... that is ...

... if \(\displaystyle gBg^{-1} = B\) for all \(\displaystyle g \in AB\) ... ... ... (1)BUT ... ...

... all elements \(\displaystyle g \in AB\) are in \(\displaystyle N_G (B)\) since \(\displaystyle AB \leq N_G (B)\) ...

... and \(\displaystyle N_G (B) = \{ g \in AB \ | \ gBg^{-1} = B \}\) ... ... ... (2)

Now, (2) implies (1) is true ...Can someone please confirm that this analysis is correct?

------------------------------------------------------------

My further question is as follows:

In the proof D&F define \(\displaystyle \phi \ : \ A \to AB/B\) by \(\displaystyle \phi (a) = aB\) ... ...

They show that \(\displaystyle \phi\) is a homomorphism and that it also is the restriction to the subgroup A of the natural homomorphism

\(\displaystyle \pi \ : \ AB \to AB/B\)

They then state:

"It is clear from the definition of \(\displaystyle AB\) that \(\displaystyle \phi\) is surjective ... ... "

Can someone please explain the thinking behind this remark?

Peter***EDIT***

In my previous post I indicated that I was unsure of why the Second Isomorphism Theorem was called the Diamond Isomorphism Theorem ... HOWEVER ... if I had taken more care and time I would have noticed that D&F explain this on pages 97-98 indicating that the reason is the diamond shape of the lattice of the subgroups of G that are involved ... ...

 

[mathbalarka]

Your analysis is correct. Actually, it'd be more clear if you generalize your statement

Claim Let $P, Q \leq G$. If $P \leq N_G(Q)$ and $Q \leq P$ then $Q$ is a normal subgroup of $P$.

Proof : Note that $N_G(Q)$, the normalizer of $Q$ in $G$, is the set of elements $g$ in $G$ such that $gQg^{-1} = Q$ for all $g \in G$. As $P \leq N_G(Q)$, $gQg^{-1} = Q$ for all $g \in P$. But that is precisely the definition of normality ($gQg^{-1} \Longrightarrow gQ = Qg$ hence the left cosets are the right cosets) thus $Q \trianglelefteq P$. $\blacksquare$

Your question is answered by subbing $P = AB$ and $Q = B \leq AB$.

"It is clear from the definition of $AB$ that $\phi$ is surjective ... ... "

Can someone please explain the thinking behind this remark?

Well. The elements of $AB/B$ are of the form $abB = aB$ for arbitrary $a \in A$ and $b \in B$. But then $aB$ is precisely $\phi(a)$. Hence every element of $AB/B$ is $\phi(a)$ for some element $a \in A$. Thus the map $\phi : A \to AB/B$ is surjective.

(Diamond Isomorphism Theorem ? why Diamond ?)

The corresponding lattice looks like a diamond, which you already seem to have figured out.

View attachment 3271​

 

[Deveno]

Some comments:

First, let's look at a "typical element" $xB$ of $AB/B$ (assuming $AB$ is indeed a group).

Since $x = ab$ for some $a \in A,b \in B$, we have:

$xB = \{abb': b' \in B\}$.

Consider the bijection $B \to B$ given by: $b' \mapsto bb'$ (here, $b$ is the fixed element of $b$ we consider writing $x = ab$, while $b'$ is allowed to vary over all of $B$).

This gives a bijection $aB \to xB$ (simply send $ab'$ to $abb'$, for every $b' \in B$), since left-multiplication by $a$ is injective.

Now, given $xb' \in xB$, we clearly have $xb' = abb' \in aB$, so $xB$ and $aB$ intersect, and thus must coincide.

So, in effect, $AB/B$ "cancels the $B$'s", giving us cosets of $A$ only.

We can only form the GROUP (and not just set of cosets) $AB/B$ when $B$ is NORMAL in $AB$. So what might that entail?

Well, for any $b' \in B$ and $ab \in AB$ we need:

$(ab)b'(ab)^{-1} = a(bb'b^{-1})a^{-1} \in B$.

Since $B$ is a subgroup of $G$, the "middle three" terms will again be in $B$.

So what we require is that $aBa^{-1} = B$, for any $a \in A$.

This is the requirement that $A \subseteq N_G(B)$.

Note that if this holds, then for any $a \in A$, and $b' \in B$, we have:

$ab'a^{-1} = b_1 \in B$, so that:

$ab' = b_1a$, that is: $AB \subseteq BA$. A similar argument shows $BA \subseteq AB$, so that $AB = BA$.

A standard theorem (which I will not prove here, but you may investigate on your own) is:

For any subgroups, $A,B$ of a group $G$, the set $AB$ is a subgroup of $G$ if and only if $AB = BA$.

One way for this to happen is for one of the subgroups to normalize the other, as we saw above.

Now, how this applies to modules is even simpler:

Since a module $M$ is always an ABELIAN group, we need not worry about normality and normalizers, submodules $A$ and $B$ ALWAYS satisfy:

$A + B = B + A$.

Thus, with no restrictions on $A$ and $B$, we may state:

$(A + B)/B \cong A/(A \cap B)$.

Note that if $A,B$ have trivial intersection, we have:

$(A+B)/B = (A \oplus B)/B \cong A/(A \cap B) = A/(\{0\}) = A$.

(one must verify that the scalar multiplication is "consistent" with this, but this is fairly straight-forward).

 

[Math Amateur]

Your analysis is correct. Actually, it'd be more clear if you generalize your statement

Claim Let $P, Q \leq G$. If $P \leq N_G(Q)$ and $Q \leq P$ then $Q$ is a normal subgroup of $P$.

Proof : Note that $N_G(Q)$, the normalizer of $Q$ in $G$, is the set of elements $g$ in $G$ such that $gQg^{-1} = Q$ for all $g \in G$. As $P \leq N_G(Q)$, $gQg^{-1} = Q$ for all $g \in P$. But that is precisely the definition of normality ($gQg^{-1} \Longrightarrow gQ = Qg$ hence the left cosets are the right cosets) thus $Q \trianglelefteq P$. $\blacksquare$

Your question is answered by subbing $P = AB$ and $Q = B \leq AB$.

Well. The elements of $AB/B$ are of the form $abB = aB$ for arbitrary $a \in A$ and $b \in B$. But then $aB$ is precisely $\phi(a)$. Hence every element of $AB/B$ is $\phi(a)$ for some element $a \in A$. Thus the map $\phi : A \to AB/B$ is surjective.
The corresponding lattice looks like a diamond, which you already seem to have figured out.

View attachment 3271​

Thanks Mathbalarka, I really appreciate your help in this matter ...

Just a minor clarification ... you write:

"The elements of $AB/B$ are of the form $abB = aB$ for arbitrary $a \in A$ and $b \in B$. "

Can you explain this statement ... surely the elements of \(\displaystyle AB/B\) are of the form \(\displaystyle abB\) and are only equal to \(\displaystyle aB\) when we have \(\displaystyle ab = a1_G = a\) (which will happen since \(\displaystyle A \subseteq AB\))

Would appreciate your help ... ...

Peter
***EDIT***

I have just been reading Deveno's latest post in this thread, and if I am right in my interpretation of his opening remarks, then he answers my question to you ...

Deveno writes:

"First, let's look at a "typical element" $xB$ of $AB/B$ (assuming $AB$ is indeed a group).

Since $x = ab$ for some $a \in A,b \in B$, we have:

$xB = \{abb': b' \in B\}$.

Consider the bijection $B \to B$ given by: $b' \mapsto bb'$ (here, $b$ is the fixed element of $b$ we consider writing $x = ab$, while $b'$ is allowed to vary over all of $B$).

This gives a bijection $aB \to xB$ (simply send $ab'$ to $abb'$, for every $b' \in B$), since left-multiplication by $a$ is injective.

Now, given $xb' \in xB$, we clearly have $xb' = abb' \in aB$, so $xB$ and $aB$ intersect, and thus must coincide.

So, in effect, $AB/B$ "cancels the $B$'s", giving us cosets of $A$ only."


Can you confirm that I am right in thinking that this answers my question to you, Mathbalarka?[Thanks again to Mathbalarka and Deveno for their help regarding the proof of the Second Isomorphism Theorem for Groups!]

 

[Math Amateur]

Some comments:

First, let's look at a "typical element" $xB$ of $AB/B$ (assuming $AB$ is indeed a group).

Since $x = ab$ for some $a \in A,b \in B$, we have:

$xB = \{abb': b' \in B\}$.

Consider the bijection $B \to B$ given by: $b' \mapsto bb'$ (here, $b$ is the fixed element of $b$ we consider writing $x = ab$, while $b'$ is allowed to vary over all of $B$).

This gives a bijection $aB \to xB$ (simply send $ab'$ to $abb'$, for every $b' \in B$), since left-multiplication by $a$ is injective.

Now, given $xb' \in xB$, we clearly have $xb' = abb' \in aB$, so $xB$ and $aB$ intersect, and thus must coincide.

So, in effect, $AB/B$ "cancels the $B$'s", giving us cosets of $A$ only.

We can only form the GROUP (and not just set of cosets) $AB/B$ when $B$ is NORMAL in $AB$. So what might that entail?

Well, for any $b' \in B$ and $ab \in AB$ we need:

$(ab)b'(ab)^{-1} = a(bb'b^{-1})a^{-1} \in B$.

Since $B$ is a subgroup of $G$, the "middle three" terms will again be in $B$.

So what we require is that $aBa^{-1} = B$, for any $a \in A$.

This is the requirement that $A \subseteq N_G(B)$.

Note that if this holds, then for any $a \in A$, and $b' \in B$, we have:

$ab'a^{-1} = b_1 \in B$, so that:

$ab' = b_1a$, that is: $AB \subseteq BA$. A similar argument shows $BA \subseteq AB$, so that $AB = BA$.

A standard theorem (which I will not prove here, but you may investigate on your own) is:

For any subgroups, $A,B$ of a group $G$, the set $AB$ is a subgroup of $G$ if and only if $AB = BA$.

One way for this to happen is for one of the subgroups to normalize the other, as we saw above.

Now, how this applies to modules is even simpler:

Since a module $M$ is always an ABELIAN group, we need not worry about normality and normalizers, submodules $A$ and $B$ ALWAYS satisfy:

$A + B = B + A$.

Thus, with no restrictions on $A$ and $B$, we may state:

$(A + B)/B \cong A/(A \cap B)$.

Note that if $A,B$ have trivial intersection, we have:

$(A+B)/B = (A \oplus B)/B \cong A/(A \cap B) = A/(\{0\}) = A$.

(one must verify that the scalar multiplication is "consistent" with this, but this is fairly straight-forward).

Thank you for a most informative and helpful post ...

Still reflecting and puzzling about your extremely interesting comments about modules ...

Struggling a little bit to see how

\(\displaystyle A + B = B + A \)

leads to

$(A + B)/B \cong A/(A \cap B)$.

with no restrictions on A and B ... ... ?Thanks again,

Peter

 

[Deveno]

Thank you for a most informative and helpful post ...

Still reflecting and puzzling about your extremely interesting comments about modules ...

Struggling a little bit to see how

\(\displaystyle A + B = B + A \)

leads to

$(A + B)/B \cong A/(A \cap B)$.

with no restrictions on A and B ... ... ?Thanks again,

Peter

In an abelian group $G$, for any subgroup $A$, we have $N_G(A) = G$, so any subgroup is contained in the normalizer of $A$.

This is because $gag^{-1} = a$.

In modules, we write the group operation as +, so $AB$ becomes $A + B$.

 

[mathbalarka]

Can you confirm that I am right in thinking that this answers my question to you, Mathbalarka?

Deveno is right about this, yes. Essentially take a look at $B/B$. The cosets are of the form $bB$ where $b \in B$. But multiplication by $g$ is a transitive action over a certain group $G$, thus $bB = B$.

This is similar to what is happening in $AB/B$. The cosets are $abB$ but multiplication by $b$ just permutes the group elements, i.e., $bB = B$. So you're left with the cosets of $A$, i.e., $aB$.

 

[mathbalarka]

Struggling a little bit to see how

\(\displaystyle A + B = B + A \)

leads to

$(A + B)/B \cong A/(A \cap B)$.

with no restrictions on A and B ... ... ?

Our restriction on $A$ and $B$ were that $A \leq N_G(B)$. But in case of modules, $G$ is an abelian group, thus $N_G(B) = G$ (any subgroup normalizes all of $G$). So the condition $A \leq N_G(B)$ is essentially equivalent to $A \leq G$ which is just superfluous.On the other hand, note that $AB = BA \iff A \leq N_G(B)$ by definition of normalizers. In case of modules, however, $G$ is abelian and thus $A$ and $B$ are also abelian, in which case $AB = BA$ automatically holds. Hence the further restriction of $A \leq N_G(B)$ is not needed.