Using the Second Isomorphism (Diamond Isomorphism) Theorem

In summary, the problem statement can only make sense if A and N are finite and if AN is a subgroup of G. The second isomorphism theorem states that AN / N is the same as A / A\cap N.
  • #1
DeldotB
117
8

Homework Statement


Good day all,

Im completely stumped on how to show this:

|AN|=(|A||N|/A intersect N|)

Here: A and N are subgroups in G and N is a normal subgroup.
I denote the order on N by |N|

Homework Equations


[/B]
Second Isomorphism Theorem

The Attempt at a Solution



Well, I know know I am supposed to use the theorem, but I have no idea where to start!
I don't seem how to relate what the theorem says to orders of sets...

Any help would be greatly appreciated!
 
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  • #2
First, to be able to use the 2nd isomorphism theorem at all, you need to prove that AN is a subgroup. Has that been given to you as a theorem? If not, can you prove it (it's quite easy)?

Next note that the problem statement only makes sense if A and N are finite.

With those out of the way, the second isomorphism theorem says:

$$AN / N\cong A/A\cap N$$
from which it follows that
$$|AN / N|=| A/A\cap N|$$
You have been asked to prove

$$\frac{|AN|}{|N|}=\frac{|A|}{|A\cap N|}$$

So if you can prove that, for any finite group ##G## and normal subgroup ##N##,
$$|G/N| =\frac{|G|}{|N|}$$
then you're almost done.

Can you prove that? Think about the size ##|gN|## of each coset ##gN##.
 
  • #3
Ahh! Thanks! I didnt realize I could just divide by |N|...I am new to quotient groups and for some reason I havnen't been able to get the hang of them yet. Thanks!

I'll use Lagranges Theorem for the last part
 
  • #4
DeldotB said:
Ahh! Thanks! I didnt realize I could just divide by |N|.
You can do that because the equation you've written is just numbers, as everything is enclosed by |.| signs. So no group theory is needed to justify the division.

Actually, the division is only justified if ##|N|\neq 0## but the proposition is untrue otherwise, since the LHS will be 0 and the RHS will be 0/0, which is undefined. So we assume that the examiner meant to stipulate that ##|N|\neq 0##.
 

Related to Using the Second Isomorphism (Diamond Isomorphism) Theorem

1. What is the Second Isomorphism (Diamond Isomorphism) Theorem?

The Second Isomorphism Theorem, also known as the Diamond Isomorphism Theorem, is a fundamental theorem in group theory that describes the relationship between a subgroup and a factor group of a larger group.

2. How does the Second Isomorphism Theorem work?

The Second Isomorphism Theorem states that if H and N are subgroups of a group G, with N being a normal subgroup, then the factor group (H⋂N)/N is isomorphic to the subgroup H/(H⋂N). This means that the factor group and the subgroup have the same structure, even though their elements may be different.

3. What is the significance of the Second Isomorphism Theorem?

The Second Isomorphism Theorem is significant because it allows us to simplify and better understand the structure of a group by breaking it down into smaller, isomorphic subgroups. It also has many applications in other areas of mathematics, such as abstract algebra and topology.

4. Can the Second Isomorphism Theorem be applied to non-abelian groups?

Yes, the Second Isomorphism Theorem can be applied to both abelian and non-abelian groups. However, for non-abelian groups, the subgroup H and normal subgroup N must have a non-trivial intersection (H⋂N≠{e}), otherwise the theorem does not hold.

5. How is the Second Isomorphism Theorem related to the First and Third Isomorphism Theorems?

The First, Second, and Third Isomorphism Theorems are all related and build upon each other. The Second Isomorphism Theorem is a special case of the Third Isomorphism Theorem, and the First Isomorphism Theorem is often used to prove the Second Isomorphism Theorem. Together, these theorems provide a powerful tool for understanding the structure of groups.

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