- #1
Bashyboy
- 1,421
- 5
Hello everyone,
I am working with an arbitrary finite group ##G##, and I am trying to prove a certain property about the order of an arbitrary element ##g \in G##. Supposedly, if we are dealing with a such a group, then ##o(g)##, which is the cardinality of the set ##| \langle g \rangle |##, is the smallest natural number ##r## such that ##g^r = e##.
Here is some of my thoughts:
If ##G## is a finite group, then there are only a finite number of elements ##g## can generate, meaning that ##\langle g \rangle## is a finite set. Let this set be
##\langle g \rangle = \{g^{-m}, g^{-m+1},...,g^{-2}, g^{-1}. g^0, g^1, g^2,...g^m \}##.
The cardinality of this set is ##2(m+1)##. If I understand the problem correctly, I need to demonstrate that ##g^{2(m+1)} = e##. Is that right?
If this is the case, could anyone provide a hint as to the next step?
I am working with an arbitrary finite group ##G##, and I am trying to prove a certain property about the order of an arbitrary element ##g \in G##. Supposedly, if we are dealing with a such a group, then ##o(g)##, which is the cardinality of the set ##| \langle g \rangle |##, is the smallest natural number ##r## such that ##g^r = e##.
Here is some of my thoughts:
If ##G## is a finite group, then there are only a finite number of elements ##g## can generate, meaning that ##\langle g \rangle## is a finite set. Let this set be
##\langle g \rangle = \{g^{-m}, g^{-m+1},...,g^{-2}, g^{-1}. g^0, g^1, g^2,...g^m \}##.
The cardinality of this set is ##2(m+1)##. If I understand the problem correctly, I need to demonstrate that ##g^{2(m+1)} = e##. Is that right?
If this is the case, could anyone provide a hint as to the next step?