# Group and subgroup ordering

1. Mar 30, 2012

### linda300

Hey,

I'm just trying to grasp ordering of groups and subgroups a little better,

I get the basics of finding the order of elements knowing the group but I have a few small questions,

If you have a group of say, order 100, what would the possible orders of an element say g^12 in the group be?

Would they just be the multiples of g^12 that have power less then 100? Like 2,3,4,5,6,7,8 ?

I read up on Lagrange's theorem but from my understanding that is related to the order of subgroups not individual group elements.

Which brings me to my next question,

Say if you have a group with two subgroups A and B, where the order of A is 120 and the order of B is say 105,

The lowest common multiple of these two numbers is 840, so would that be the order of the group? Since both the orders of the subgroups need to be a multiple of the order of the group?

But if you take the intersection between the two subgroups A and B, what would be the possible orders of the intersection?

Would one of them be 120-105 = 15? If all of the elements in B are also in A.

Thanks,
Linda

2. Mar 30, 2012

### jgens

I have no idea what you mean by this. Can you be more clear?

It does relate to the order of an individual group element. If $G$ is a group and $x \in G$, then $|\langle x \rangle| = |x|$; that is, the order of each group element must divide the order of the group.

No. But it means that the order of the group is an integer multiple of 840.

In general this question is difficult to answer. You know that if $H_1,H_2 \leq G$, then $H_1 \cap H_2 \leq G$. So in particular the order of $H_1 \cap H_2$ necessarily divides the order of $G$. To say much more than that usually requires knowing more information about $G$ and $H_1,H_2$.

3. Mar 30, 2012

### linda300

Thanks jgens!

If you let G be a group of order 100, what are the possibilities for the order of g12?

So the order of g12 has to divide 100, the only multiples of 12 which give a number less then 100 are 2 and 5.

Would that mean the possibilities for the order of g12 are 2 and 5?

Is it correct to assume there are no powers of g in G which are greater then 100?

Or if you don't assume that would the possibilities be order 2, 5, 10, 20, 25, 50, just the divisors of 100

For the intersection question,

Depending on the size of G,

Could you say that the possibilities of the order of the intersection between H1 and H2 are multiplies of 840 which are less then the order of G?

So if the order of G was 2520 the possible orders of the intersection are 840 and 1680?

Last edited: Mar 30, 2012
4. Mar 31, 2012

### jgens

If $|g| = 100$, then it follows that $|g^{12}| = \frac{\mathrm{lcm}(12,100)}{12} = 25$. If $|g| = 50$, then it follows that $|g^{12}|= \frac{\mathrm{lcm}(12,100)}{12} = 25$. If $|g| = 25$, then it follows that $|g^{12}| = \frac{\mathrm{lcm}(12,25)}{12} = 25$. If $|g| = 20$, then it follows that $|g^{12}| = \frac{\mathrm{lcm}(12,20)}{12} = 5$. If $|g| = 10$, then it follows that $|g^{12}| = \frac{\mathrm{lcm}(12,10)}{12} = 5$. If $|g| = 5$, then it follows that $|g^{12}| = \frac{\mathrm{lcm}(12,5)}{12} = 5$. If $|g| = 4$, then it follows that $|g^{12}| = 1$. If $|g| = 2$, then it follows that $|g^{12}| = 1$. If $|g| = 1$, then it follows that $|g^{12}| = 1$. So the possible orders are 1, 5, 25.

Asking for the possible order of the subgroup intersection is a much more difficult question. Unless you have a specific group in mind (not just an order), this question is very difficult.