1. ## element

Suppose G has order 33. Show that it contains an element of order 3.

So G can contains elements of order 3,11, and 33. If G contains an element of order 33, then this will generate the whole group. So suppose G contains no element of order 33. Then every non-identity element of G has order 3 or 11. If every non-identity element has order 11, then the set $\displaystyle H = \{e, a_{1}, \dots, \underbrace{a_{11}, a_{1}a_{2}, \dots, a_{1}a_{11} \dots}_{11^{11} \ \text{terms}} \}$ is not a subgroup of G. E.g. there are no subgroups of G so that $\displaystyle |H| > |G|$. Thus there must be some element of order 3.

2. So you have to show that there is a subgroup of some order that does not divide 33?

3. Originally Posted by Sampras
Suppose G has order 33. Show that it contains an element of order 3.

So G can contains elements of order 3,11, and 33. If G contains an element of order 33, then this will generate the whole group. So suppose G contains no element of order 33.
Why? Are you saying that if G contains an element of order 33 then it is obvious that it contains an element of order 3? I think you should show that as well.

Then every non-identity element of G has order 3 or 11. If every non-identity element has order 11, then the set $\displaystyle H = \{e, a_{1}, \dots, \underbrace{a_{11}, a_{1}a_{2}, \dots, a_{1}a_{11} \dots}_{11^{11} \ \text{terms}} \}$ is not a subgroup of G. E.g. there are no subgroups of G so that $\displaystyle |H| > |G|$. Thus there must be some element of order 3.
Excuse me but what do you mean by $\displaystyle a_{1}$, $\displaystyle a_{11}$, etc.?

4. Originally Posted by HallsofIvy
Why? Are you saying that if G contains an element of order 33 then it is obvious that it contains an element of order 3? I think you should show that as well.

Excuse me but what do you mean by $\displaystyle a_{1}$, $\displaystyle a_{11}$, etc.?
No I'm trying to show that if every non-idenity element had order 11 then this would be a contradiction. Because we would have a subgroup of some order that does not divide 33.

5. What exactly can you use buddy? Can you use Sylow? I'd assume not or you wouldn't be asking.

6. Originally Posted by Drexel28
What exactly can you use buddy? Can you use Sylow? I'd assume not or you wouldn't be asking.
No but I think I figured it out. Consider $\displaystyle H_1 = \{e, a^2, \dots , a^{10} \}$, $\displaystyle H_2 = \{e, b, \dots, b^{10} \}$ and $\displaystyle H_3 = \{e, c, \dots, c^{10} \}$ where $\displaystyle b \notin G$. Then $\displaystyle a^{i} = b^{j}$ for some i and j. Or $\displaystyle (a^i)^{l} = (b^{j})^{l}$ for some $\displaystyle l$. Contradiction.