Question

Need help with the following:
-Show that a cyclic group of order 22 has one element of order 1 and 10 elements of order 11.
-Show that a noncyclic group of order 22 has 21 elements each of order 2 or 11.

Answer 1

Well every group always has exactly one element of order 1, so that part is easy. Now let's find the number of elements of order 11 in the cyclic of order 22.

Answer 2

Let \(g\) generate the group. Then, notice that \((g^{2k})^{11}\) is the identity for any integer \(k\). Furthermore, if \(1\le k\le 10\), then no lower power of \(g\) will result in the identity. Thus, there are 10 elements of order 11.

Answer 3

That make sense?

Answer 4

yes but what is k again?

Answer 5

wait nvm I see that you said any integer k

Answer 6

How does it work for noncyclic groups?

Answer 7

For non-cyclic groups, you still have the single element \(e\) with order 1. Now, can you list out all the possible orders of an element in a group of order 22?

Answer 8

so for order 2: (g^k)^2, 1<=k<=21 (Therefore 21 elements of 2)
order 11: (g^k)^11, 1<=k<=21 (Therefore 21 elements of order 11)
This is probably wrong..

Answer 9

Sorry for the absence there. To cut to the chase, that is wrong. But you're in luck because we're using simpler ideas for the second.

May I presume that you've covered Lagrange's Theorem? I.e., the order of subgroups divides the order of the group?

Answer 10

Yes i did but my professor kind of breezed through it without much explanation.

Answer 11

That's an important part to the answer to the second question. So by applying Lagrange's Theorem, what are the possible orders for an element in a group of order 22?

Answer 12

For Lagrange's theorem, in my notes I have this: Let G be a finite group and H a subgroup of G. Then the order of H, |H| the number of elements in H, divides the order of G.

Answer 13

So for order 2, There is |21| elements in order 2. And it divides whatever G is?

Answer 14

Back up to Lagranges Theorem. If |G|=22, what numbers are possible for |H|?