## The Lattice Isomorphism Theorem

Prove all parts of the Lattice Isomorphism Theorem.

Let be a group and a normal subgroup, with the natural projection . Let denote the set of all subgroups of containing and the set of all subgroups of , and define by . Let . Then we have the following.

- is a bijection. (Injective) Suppose ; then . Now let . We have for some , so that . Then , and we have . Similarly, if then . Thus , and is injective. (Surjective) Let ; we saw previously that the preimage of a subgroup under a group homomorphism is a subgroup, so that there exists some such that .
- if and only if . If , then , so . Thus . Suppose and let . Now for some , so that . Then , hence .
- If , then . If , then by the previous part, we have partitions and and every equivalence class has the form or , respectively. Define by . (Well defined) If , then , so , hence . (Injective) Suppose . Then , and we have for some . Now , so that . Thus . (Surjective) is surjective because is surjective. Thus is a bijection and the conclusion follows.
- . Let . Then for some , so that for some . Then , where . Thus . If , then for some . Now , and for each . Thus for some , so that .
- . If , then for some . Now , so , and similarly , so . If , then for some and . Now , so that , hence . Thus .
- is normal if and only if is normal. Suppose is normal, and let . Because is surjective, we have for some . Now , so that is normal. Now suppose is normal and let . We have . Thus if , we have for some , so . Thus is normal.

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## Comments

error: 1,2. implies , so that . (not but )

3. should be . And the same errors are in there. For example, implies

5. Similarly, .

6. should be . Also the same error: From , we get for some , so that . (not but \in N$

my additional explanation: 1.(surjective) The surjectivity of is important. Let be any subgroup of . Define . Then since is surjective. Obviously so .

6. Just using the equivalent condition is also straightforward. is normal iff for all iff for all iff is normal in .

question: In 4, is it true that ? Isn’t the proof necessary? Actually, I think that we only need , which is quite straightforward.

ps. Is there an easy way to type your image-filed-latex-code?

I was halfway though fixing these errors when I realized I wrote this proof using the ‘right’ congruence induced by rather than the usual ‘left’ congruence, hence the consistent backwardness. I went ahead and changed the proof to use the left coset notation. (Given any subset , we can define two relations on using , one ‘left’ and one ‘right’, by iff and iff . These are equivalences if and only if is a subgroup (a pleasant fact to check) and are congruences (i.e. the naturally induced operator on classes is well defined) if and only if is normal. The two relations are dual, so it doesn’t matter which one we use so long as we’re consistent.)

As far as easy typing, I find that the best way to get faster at typing latex is to type lots of latex. đź™‚

ps2. Is there any way to edit my comments? Or is there anywhere to check the result of my comments? My elaborate comment above has some typos, and I’ll be glad if you fix them.

Unfortunately there is no way to preview comments as far as I know- although that would be an incredibly useful feature.