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**jjbrian** In Z/6Z define [a] * [b] =[a+b-1]. Is that associative? If so, does this operation make Z/6Z into a group? If its not in the group explain why?

Attempt:

([a] * [b]) * [c] = [a + b - 1] * [c] = [a + b + c - 2]

[a] * ([b] * [c]) = [a] * [b + c - 1] = [a + b + c - 2] = ([a] * [b]) * [c]

Thus, the operation is associative. Now, we just need to check identity and inverses. What's the identity? Well, we need some [e] such that:

[a] * [e] = [e] * [a] = [a]

Thus we need:

[a] * [e] = [a + e - 1] = [a]

for all a. That is, we need:

a + e - 1 = a (mod 6)

e - 1 = 0 (mod 6)

e = 1 (mod 6)

So [1] is an identity (and thus the unique identity). So, now can we compute an inverse for each element? Suppose we have some [a], and it has an inverse [b]. Then:

[a] * [b] = [1]

[a + b - 1] = [1]

a + b - 1 = 1 (mod 6)

a + b = 2 (mod 6)

b = 2 - a (mod 6)

Thus, if an inverse of [a] exists, it must be [2 - a]. We can check that [2 - a] is the inverse for each [a]:

[2 - a] * [a] = [a] * [2 - a] = [a + 2 - a - 1] = [1]

Thus, it works, and we have a group.

2.In the dihedral group D3, are there any repetitions in the set[ I, r, r^2, t, tr, tr^2}of rigid motions?

No. (I dont know, just a guess).

Why "guess"? Prove it! Of course, we must assume $\displaystyle ord(r)=3\,,\,ord(t)=2$ , and then $\displaystyle t=r$ impossible, $\displaystyle t=tr\Longrightarrow r=1$ , impossible again, and etc.

Tonio

Can someone check my work for these two problems?