2

I have a quick question about the first isomorphism theorem. In our class the isomorphism theorem is broken up into several parts, but the first part is as such:

If $f$ is a factor map from $G \to G/K$ (where $K$ is a normal subgroup) then $f$ defines a one to one correspondence between all subgroups $H$ of $G$ containing $K$ and all subgroups of the factor group $G/K$.

I am confused as to what this theorem is saying, is it saying that if we consider the set of all subgroups such that $K$ is in $H$ that that will be isomorphic to the set of all subgroups $H$ is in $G/K$? Or is it saying that each subgroup that contains $K$ is isomorphic to each subgroup contained in $G/K$? The second possibility does not seem correct to me, but the first does not seem incredibly useful in the context of the theorem. I guess I just wanted to know if there was a less wordy and more symbol driven way of writing this statement.

Thank you in advance!

pjs36
  • 17,979
MathStudent
  • 31
  • 1
    This just says that there is an order preserving correspondence. Or, there is a bijection (just on the level of functions between sets) between the set of groups of $G$ containing $K$ and the subgroups of $G/K$. Usually this is called the correspondence or Fourth isomorphism theorem. – Eoin Apr 01 '15 at 03:01

3 Answers3

3

One of the ways to think about this theorem is the following: Since $K$ is normal in $G$, the quotient $G/K$ is a group. And, of course, an important question to ask about any group is: What are its subgroups? This theorem answers that question in a nice and complete way.

Ex. What are the subgroups of $\mathbb{Z}/n\mathbb{Z}$ for some fixed $n$? According to our theorem, they correspond precisely to those subgroups of $\mathbb{Z}$ that contain $n\mathbb{Z}$. These are the groups $k\mathbb{Z}$, where $k$ divides $n$. Projecting these down into the factor group, we see that the subgroups of $\mathbb{Z}/n\mathbb{Z}$ are all of the form $k\mathbb{Z}/n\mathbb{Z}$, for $k \, | \, n$.

PeterJL
  • 1,160
2

Say $N\trianglelefteq G$ and let $\pi:G\to G/N$ be the projection. The theorem states there is a bijection

$$\{H\,|~N\subseteq H\subseteq G\}\quad\longleftrightarrow\quad \{K\,|~K\subseteq G/N\}. \quad$$

The mutually inverse maps are $H\mapsto\pi(H)$ and $K\mapsto\pi^{-1}(K)$ (the inverse image).

Moreover, the two sets above are ordered by inclusion (in fact they are lattices), and both maps preserve the ordering, so this bijection is actually an isomorphism of lattices.

anon
  • 151,657
  • So if order to prove this I would need to prove that f^-1(f(H))=H and that f(f^-1(K))=K, is that correct? So, as a general sketch, I could prove that H must be carried to H/N, which must be carried back to H? – MathStudent Apr 01 '15 at 03:23
  • And then do the corresponding inverse of that process? – MathStudent Apr 01 '15 at 03:23
  • @MathStudent Yes to your first comment. Not sure I understand your second comment, since we're already discussing both directions of the correspondence. – anon Apr 01 '15 at 13:24
0

the basic idea, shed of most of its complexities, can be appreciated by analogy with the factor structure of the positive integers. just for illustration purposes the analogy is that an integer is a "group" and that $m$ divides $n$ stands for $M$ is a subgroup of $N$

let G be the number $120$ and consider the set $L=\{120, 60, 40, 30, 20, 15, 12\}$. each number in $L$ divides $G$. however, if you divide each number by $20$ you get $L' = \{ 6,3,2,\frac32,1, \frac34, \frac35 \}$

so here your normal subgroup $K$ "corresponds" to the number $20$. you notice that the numbers divisible by $20$ give integers, whereas the fractions arise from the two numbers which are not divisible by $20$. similarly the subgroups which contain the normal subgroup $K$ remain subgroups of the quotient group $G/K$, whereas the other subgroups do not survive.

David Holden
  • 18,040