Characterizing subgroups of a cyclic group

  • #1
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Homework Statement


Show that for every subgroup ##H## of cyclic group ##G##, ##H = \langle g^{\frac{|G|}{|H|}}\rangle##.

Homework Equations




The Attempt at a Solution


At the moment the most I can see is that ##|H| = |\langle g^{\frac{|G|}{|H|}}\rangle|##. This is because if ##(g^{\frac{|G|}{|H|}})^p = e## for some value of ##p < |H|##, we would have that ##g^q = e## for some ##q < |G|##, which is a contradiction.

But I'm not seeing why they must be the same subgroup. I was going to try to prove that if two subgroups of cyclic group have the same order, then they must be the same group. If I did this I think I would first need to prove that every subgroup of a cyclic group is cyclic.
 
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Answers and Replies

  • #2
14,389
11,708

Homework Statement


Show that for every subgroup ##H## of cyclic group ##G##, ##H = \langle g^{\frac{|G|}{|H|}}\rangle##.

Homework Equations




The Attempt at a Solution


At the moment the most I can see is that ##|H| = |\langle g^{\frac{|G|}{|H|}}\rangle|##. This is because if ##(g^{\frac{|G|}{|H|}})^p = e## for some value of ##p < |H|##, we would have that ##g^q = e## for some ##q < |G|##, which is a contradiction.

But I'm not seeing why they must be the same subgroup. I was going to try to prove that if two subgroups of cyclic group have the same order, then they must be the same group, but I think that's basically what I'm already trying to prove
Haven't you done this already?
Don't they not generate the group because they have a different order than the generator ##g##, since ##|g^k| = \frac{n}{\operatorname{gcd} (k,n)}##?
With ##|G|=n## and ##k=|H|## we have ##k\,|\,n##, say ##n=k\cdot m## and we are interested in the order of ##g^m##. Now apply your formula.
 
  • #3
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Haven't you done this already?

With ##|G|=n## and ##k=|H|## we have ##k\,|\,n##, say ##n=k\cdot m## and we are interested in the order of ##g^m##. Now apply your formula.
Well what we have is that ##|\langle g^{\frac{|G|}{|H|}} \rangle| = |g^{\frac{n}{k}}| = |g^m| = \frac{n}{\gcd (m,n)} = \frac{mk}{\gcd (m,mk)}= \frac{mk}{m} = k = |H|##. So now I have that the orders are equal. But how I should I go about showing that they are the same group?
 
  • #4
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11,708
Well what we have is that ##|\langle g^{\frac{|G|}{|H|}} \rangle| = |g^{\frac{n}{k}}| = |g^m| = \frac{n}{\gcd (m,n)} = \frac{mk}{\gcd (m,mk)}= \frac{mk}{m} = k = |H|##. So now I have that the orders are equal. But how I should I go about showing that they are the same group?
Let ##H':=\langle \,g^m\,|\,g^n=1\, , \,n=mk \,\rangle## and ##g^l \in H##. Now what does the last property mean, if we only know that ##|H|=k\,?##
 
  • #5
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Let ##H':=\langle \,g^m\,|\,g^n=1\, , \,n=mk \,\rangle## and ##g^l \in H##. Now what does the last property mean, if we only know that ##|H|=k\,?##
All I can see is that if ##g^l \in H##, then ##g^{lk} = 1##, and so ##n ~|~ lk##.
 
  • #7
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44
Or ##lk=a\cdot n = amk## ...
Is the point that since ##l = am##, we then know that ##g^l = (g^m)^a##, so ##g^l \in H'##?
 
  • #8
14,389
11,708
Is the point that since ##l = am##, we then know that ##g^l = (g^m)^a##, so ##g^l \in H'##?
Yes, and this is true for all elements of ##H## since we didn't set any conditions on the element - don't forget that ##|H'|=k=|H|##.
 
  • #9
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44
Yes, and this is true for all elements of ##H## since we didn't set any conditions on the element - don't forget that ##|H'|=k=|H|##.
So since ##H \le H'## and ##|H| = |H'|##, we have ##H = H'##?

Could you explain a little bit why ##\langle g^{\frac{n}{k}}\rangle = \langle g^m~|~g^n=1, n = mk \rangle##? And also what the notation on the RHS means exactly.
 
  • #10
14,389
11,708
So since ##H \le H'## and ##|H| = |H'|##, we have ##H = H'##?
Yes.
Could you explain a little bit why ##\langle g^{\frac{n}{k}}\rangle = \langle g^m~|~g^n=1, n = mk \rangle##? And also what the notation on the RHS means exactly.
##\langle g^{\frac{n}{k}}\rangle = \langle g^m \rangle## by the definitions of ##k,m,n##

However, this notation is strictly seen not the group we deal with, because here ##g## would be of infinite order as the order isn't mentioned. The correct way to write this group is ##\langle g^{\frac{n}{k}}\rangle = \langle g^m~|~g^n=1, n = mk \rangle##, because it lists all relevant data. In general the notation is
$$
G = \langle \, g_1, \ldots ,g_n \,|\, r_1,\ldots ,r_m \,\rangle \text{ or } G = \langle \, g_1, \ldots ,g_n \,|\, r_1=\ldots =r_m=1 \,\rangle
$$
where the ##g_i## are generators of the group, possibly infinitely many, and ##r_j## are the relations, i.e. words built over the alphabet ##g_i## which equal ##1##. The relation, in our case only one, namely ##r_1=g^n=1##, makes the group cyclic and finite. ##n=mk## could have been omitted so that ##H=\langle \,g^m\,|\,g^{mk}\,\rangle## fits the just defined notation. ##r_j=1## is usually omitted, as the position behind ##"|"## indicates that it is a relation, so they are simply listed.
 

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