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Abstract Algebra Question
- Thread starter 74912dan
- Start date
\(\displaystyle e,g^k, g^{2k},...,g^{(n-1)k}\) are distinct powers of little g and they all belong to J (why?).
Since there are only n distinct powers of g (in H) to begin with, this is all of them. So \(\displaystyle g^{mk}=g\) for some \(\displaystyle m\in\{0,...,n-1\}\).
Since there are only n distinct powers of g (in H) to begin with, this is all of them. So \(\displaystyle g^{mk}=g\) for some \(\displaystyle m\in\{0,...,n-1\}\).
I am just still confused as to what it means for g^k to be an element of J when we know nothing about what kinds of elements are in J.
We know the elements in J are a subset of the elements of H, and that is all we know. To see that relative primarily is necessary:
Let H = <g> be the cyclic group of order 6 and J be the subgroup <g^2>. Now, g^4 is an element of J but g does not belong to J.
so we know g and g^k are subgroups of H. only other elements with powers relatively prime to k can be in H and J?
Huh? If k is relatively prime to |g| then <g> = <g^k>. This is immediate from |g|=|g^k|=n. Your statement in the original post lead me to believe you already knew this.
Yes I know that <g> = <g^k> and that |g|=|g^k|=n. I just can't figure out why just knowing that fact makes it so the only way g^k can be in J is if g is J.
That is not what the problem is asking. If g belongs to J then g^k will always belong to J, because J is a subgroup.
This question is doing a partial converse to the above statement. It is asking IF g^k belongs to the subgroup J for some k which is relatively prime to |g|, then it must contain g (and hence every power of g). My (why?) above was pointing at the fact that J is a subgroup, and so if it contains any element, it must contain any power of that element.
Now, with that in mind, SUPPOSE g^k belongs to J. Then <g^k> (all powers of g^k) is a subgroup of J. But <g>=<g^k> since (k, |g|)=1, so g belongs to J.
Ohhhh. I kept over-looking the fact <g> = <g^k> and focused on |g|=|g^k|=n. The question makes a lot more sense when thinking of it as <g^k> being a subgroup of J when g^k is an element of J. Anyways thanks a lot for the help, hopefully I didn't annoy you too much with my dumb questions that kept missing the obvious connections. Thanks again you helped a ton. :-D