is anyone certain of what question 4 means?

(4) Which elements commute with every other element?

Melissa & everyone else,

If an element, say $a\in G$, commutes with every other element, then this means $ax=xa$ for all $x\in G$. It's like being abelian, but you're only checking whether an individual element commutes with everything else. If the group is already abelian, then ALL the elements commutes with all the other elements. However, in a nonabelian group, some elements may still commute with every other element. Look at the group table and see if there are any rows that are identical to the corresponding column. Do you see why this is checking what I'm asking? Maybe the answer is "none." There is always *at least one* element that commutes with everything.

I hope that helps.

thanks Dana we were lost at that question.

I don't know if I was sleeping in class or what but how do we know when to either add or multiply in a cayley table? This is making me have a difficult time trying to interpret the cayley table for $H$ in the sage lab.

Shaun, the answer is that it doesn't matter what the operation is. The table tells you how to combine any two elements of the group regardless of the operation. Since I alluded to what each of these groups is, you can figure out what the operation is, but you do not need that information to answer any of the questions. You can get everything that you need from the table.

By the way, the `x0`, `x1`, etc notation is short for $x_0, x_1$, etc.

I was wondering if anyone could explain to me how to figure out if theire is a non trivial subgroup, I think I'm just stuck on how to read the graph it gave me.

Now that we've covered a little bit of chapter 3, the easiest thing to do is to find the cyclic subgroup generated by something. Pick any of the non-identity elements. Find the cyclic subgroup generated by that element. If you get the whole group, try a different element. Hint: you should end up with a subgroup of order 2 or 3.

I'll accept Sage lab 2 anytime up until midnight tonight if you are still working on it.