How many different groups are there with 4 elements?

How many groups, up to isomorphism, can we find that have order (cardinality) 4?

I show that there are only two groups, up to isomorphism, with 4 elements. One of them is the group of integers modulo 4, and the other is the Klein 4 group.

[graf_paper]

Wow, this paired so nicely with the independent investigation I was doing as I worked by way through T. Judson's Abstract algebra - realizing the connection between latin squares and groups was really nice and learning that we can look at the subgroup structure to get a handle on wether two groups are isomorphic or not. Very nice.

[aryanpatel]

That what I was searching for at 3am morning. Thanks

[bowlofsoba]

Thank you so much! This is the video I am looking for. It would be great if you can talk more why A C D are isomorphic, still struggling there.

[anantkumar]

Is there an equation or a recurrence relation that would give us the number of groups that are distinct, up to isomorphism with the general order 'n' ?

[wiggles]

At 5:30 when we have our 2 possible groups, A and B, don't we need to check that they are associative? My memory is not exact, but I used this Latin square method of finding all the groups of order 5, and I believe I ended up with a Latin square describing an operation that was not associative, and so the Latin square actually did not describe a group operation.

[prbprb]

By the way, the associative property comes as a byproduct of the other desired properties in this example presented here. A question in my own mind, is why not more generally study loops and not groups? What is so dear to the soul of having an associative table?

[egenverdi]

I can't see the link to the proof you wanted to link to at 2:08

[justfocusonurdreams]

How to show G={1, -1, i, -i}is abelian

[fortnitechad]

This doesn't even make sense. You can set up any of the 4 in this video and see that there are 4! * 2 permutations that follow her rules because you swap rows and collumns to make a new table that still satisfies the axoims.