1. ## Group Proof Question

Question:

Could anyone show me how to prove this question? So, I know that the order of every element is the smallest integer n such that $\displaystyle g^n=e$, in this case $\displaystyle (g^2)^n = g^3 = e$. I also know that the order of each element of G are arithmetically related to the order of the group, they divide the order of the group. So the order of an element and the order of the group must both be either odd or even (but I don't know how we can prove that they are related). Any help is really appreciated.

2. Originally Posted by demode
Question:

Could anyone show me how to prove this question? So, I know that the order of every element is the smallest integer n such that $\displaystyle g^n=e$, in this case $\displaystyle (g^2)^n = g^3 = e$. I also know that the order of each element of G are arithmetically related to the order of the group, they divide the order of the group. So the order of an element and the order of the group must both be either odd or even (but I don't know how we can prove that they are related). Any help is really appreciated.
Use the hint. Essentially, you have that $\displaystyle g^3=1$ and $\displaystyle g^2=g^{-1}$ with $\displaystyle g^{-3}=1$. Do you see why this is? Therefore, for every $\displaystyle g\neq 1$ such that $\displaystyle g^3=1$ there is a unique element $\displaystyle g^{-1}$ which also has order 3 and is non-trivial. Thus, there are an even number of elements of order 3. Do you understand why?

Now, can you think of the only element in your group which does not have order 3 but which the equation $\displaystyle g^3=1$ still holds? This element does not have a pair, and so you have 2n+1 elements which are solutions to this equation.

3. Originally Posted by demode
Question:

Could anyone show me how to prove this question? So, I know that the order of every element is the smallest integer n such that $\displaystyle g^n=e$, in this case $\displaystyle (g^2)^n = g^3 = e$. I also know that the order of each element of G are arithmetically related to the order of the group, they divide the order of the group. So the order of an element and the order of the group must both be either odd or even (but I don't know how we can prove that they are related). Any help is really appreciated.

Hints:

1) If the order of $\displaystyle g$ is 3 then also the order of $\displaystyle g^2$ is 3

2) Now pair up $\displaystyle \{g,g^2\}$ for each and every element of order 3 in the group

3) The number of elements of order 3 is the number of pairs you got above, and this number clearly is even (why?)

4) Finally, to the number in (3) you must add 1 to obtain the total number of elements that fulfill $\displaystyle x^3=1$

Tonio

4. Originally Posted by tonio
Hints:
3) The number of elements of order 3 is the number of pairs you got above, and this number clearly is even (why?)
Why is the number of pairs $\displaystyle \{g,g^2\}$ even if we don't know how many elements there are? I'm a bit confused...

5. Originally Posted by demode
Why is the number of pairs $\displaystyle \{g,g^2\}$ even if we don't know how many elements there are? I'm a bit confused...

Because $\displaystyle g\neq g^2$ , lest $\displaystyle g=1$ , and also because the two elements $\displaystyle g\,,\,g^2$ appear in two sets: $\displaystyle \{g,g^2\}\,,\,\,\{g^2,g=(g^2)^2=g^4\}$ , so

after you divide the number of pairs by two (to eliminate this repetition) you STILL have an even number of elements in

the remaining sets with a pair of (different, of course) elements...

Tonio