Group and subgroup ordering

[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

 

[jgens]

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

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

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

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.

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?

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

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

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 $H_1,H_2$.

 

[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?

 

[jgens]

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

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.

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?

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.

 

[blue_raver22]

Hi Linda,

Ordering of groups and subgroups can be a bit tricky, but I'll try to clarify some of your questions.

Firstly, the order of an element g^12 in a group of order 100 would depend on the specific group and the properties of g. It is not necessarily limited to just the multiples of g^12 with powers less than 100. For example, if g is a cyclic group of order 100, then g^12 would have an order of 10. However, if g is a non-cyclic group, then the order of g^12 could be any factor of 100.

Lagrange's theorem does relate to the order of subgroups, but it can also be used to determine the order of individual group elements. For example, if the order of a group is 100 and an element g has an order of 10, then by Lagrange's theorem, the order of g must be a factor of 100.

In your second question, the order of the group would not necessarily be the lowest common multiple of the orders of the subgroups A and B. It could be a multiple of the lowest common multiple, but it could also be a different number altogether. The order of a group is determined by the properties and operations of the group, not just the orders of its subgroups.

Finally, the order of the intersection between two subgroups A and B would depend on the specific elements in A and B. In general, the order of the intersection would be a factor of both the orders of A and B, but it could also be a smaller number depending on the specific elements in the intersection. In your example, the order of the intersection could be 15 if all elements in B are also in A, but it could also be a smaller number if there are elements in B that are not in A.

I hope this helps clarify some of your questions. If you have any further questions, please feel free to ask. Good luck with your studies!

Best,