Group and subgroup ordering

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 let G be a group of order 100, what are the possibilities for the order of g^{12}?

Would they just be divisors of 100? 2, 5 , 10, 20, 25 and 50?

Or would the power of g be restricted to below 100, giving orders of only 2 and 5?

Another question I have regards the order of an intersection of subgroups,

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,

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

Would it just be multiples and divisors of 840 which are less then the order of the group?

Thanks in advance

Re: Group and subgroup ordering

it's hard to say what the order of g^{12} might be, as we have no way of knowing just from |G| = 100, what |g| even is. the best we can do is just that g^{12} is some divisor of 100. if G is cyclic, we can say more, because cyclic groups are more "predictable". if G is merely just abelian that is somewhat helpful, because abelian groups are direct products of cyclic groups.

as to your other question, clearly A∩B has to have an order which is a common divisor of |A| and |B|, which leaves us with 1,3,5, or 15.

Re: Group and subgroup ordering

Quote:

Originally Posted by

**Angela11** If you let G be a group of order 100, what are the possibilities for the order of g^{12}?

Would they just be divisors of 100? 2, 5 , 10, 20, 25 and 50?

Or would the power of g be restricted to below 100, giving orders of only 2 and 5?

The order of any element of a finite group must divide the order of the group. The fact that it’s a power of another element of the group is immaterial. However, note that any element to the power of the order of the group is always the identity. Since $\displaystyle \left(g^{12}\right)^{25}=g^{300}=\left(g^{100} \right)^3=1_G,$ the order $\displaystyle g^{12}$ does not have to be greater than $\displaystyle 25.$ Hence the order of $\displaystyle g^{12}$ can be $\displaystyle 1,$ $\displaystyle 2,$ $\displaystyle 5,$ $\displaystyle 10,$ $\displaystyle 20,$ or $\displaystyle 25.$ (You left out $\displaystyle 1$ but note that $\displaystyle g^{12}$ can have order $\displaystyle 1$ if $\displaystyle g$ itself is the identity or has order $\displaystyle 2).$

Quote:

Originally Posted by

**Angela11** Another question I have regards the order of an intersection of subgroups,

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,

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

Would it just be multiples and divisors of 840 which are less then the order of the group?

The intersection is a smaller group than the two original groups, so you should be looking at the greatest common divisor rather than least common multiple. Lagrange’s theorem still applies to subgroups of a subgroup: the order of $\displaystyle A\cap B$ must divide both the order of $\displaystyle A$ and the order of $\displaystyle B;$ hence it must divide their GCD, which is $\displaystyle 15.$ So the possible orders of $\displaystyle A\cap B$ are $\displaystyle 1,$ $\displaystyle 3,$ $\displaystyle 5,$ and $\displaystyle 15.$

Re: Group and subgroup ordering

Thanks a lot for your replies guys,

So just to recap,

The power of the element does not matter with regards to the power, so the possible orders of g^{12} are just the dividers of 100?

1,2,5,10,20,25,100

So really if you were asked a question like this, it doesn't matter what element of the group you are given, with the information you have about the order of the group you can find the possibilities of the order of its elements?

so g^{21} would have the same possible orders as g^{12}?

Re: Group and subgroup ordering

Quote:

Originally Posted by

**Angela11** Thanks a lot for your replies guys,

So just to recap,

The power of the element does not matter with regards to the power, so the possible orders of g^{12} are just the dividers of 100?

1,2,5,10,20,25,100

Sorry, I was editing my post and didn’t check if you were posting at the same time. The order of $\displaystyle g^{12}$ does not have to be greater than $\displaystyle 25$ since $\displaystyle \left(g^{12}\right)^{25}=g^{300}=\left(g^{100} \right)^3=1_G$ (any element to the power of the order of the group is the identity).

Quote:

Originally Posted by

**Angela11** so g^{21} would have the same possible orders as g^{12}?

No, this is slightly different. Since $\displaystyle 21$ and $\displaystyle 100$ are coprime, the order of $\displaystyle g^{21}$ can be any positive divisor of $\displaystyle 100,$ including $\displaystyle 50$ and $\displaystyle 100$ this time.

Re: Group and subgroup ordering

Ah I see!

Thanks so much, this really helped!

Re: Group and subgroup ordering

Hey I was just going over this question again and noticed that 4 isn't mentioned, isn't 4 a divider of 100? 4x25

So 4 would be a possible order of g^12 as well right?

Re: Group and subgroup ordering

yes, 4 is a possibility of a general element of a group of order 100. but can g^{12} have order 4?

no. suppose it did: then (g^{12})^{4} = g^{48} = e.

this implies that |g| divides 48, and 100, so |g| divides 4 = gcd(48,100), thus |g| = 1,2 or 4.

|g| = 1 implies g = e, so g^{12} = e^{12} = e, which has order 1.

|g| = 2 implies g^{12} = (g^{2})^{6} = e^{6} = e, which has order 1.

|g| = 4 implies g^{12} = (g^{4})^{3} = e^{3} = e, which again had order 1.

the fact that (g^{12})^{25} = g^{300} = (g^{100})^{3} = e^{3} = e,

means that |g^{12}| must divide 25, so |g^{12}| can only be 1,5 or 25.

this is a consequence of the following theroem:

if g in G has order k, and g^{n} = e for n > 0, then k divides n.

proof:

clearly n is greater than k (since k is the least positive integer for which g^{k} = e).

so we can write n = qk + r, where 0 ≤ r < k. since g^{n} = e, we have:

e = g^{n} = g^{(qk+r)} = (g^{(qk)})(g^{r}) = (g^{k})^{q}(g^{r})

= (e^{q})(g^{r}) = eg^{r} = g^{r}.

since the order of g is k, we cannot have 0 < r < k, which forces r = 0. thus n = qk, which means that k divides n.