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seen Dec 17 at 6:16

Feb
6
comment Finding the order of elements in a group?
Alright, so I picked up where you left off, and I found the following. $[4]^2 = [16]$, so $n = 2$ does not work. However, $[4]^{11} = [1]$, so $n = 11$ and since $11$ is least, then the order of $[4]$ is 11. Am I correct?
Feb
6
accepted Finding the order of elements in a group?
Feb
6
comment Finding the order of elements in a group?
Thank you so much for the clarification! I think I understand this problem now.
Feb
6
comment Finding the order of elements in a group?
Yes, that is the exact question. Sorry for any confusion, I thought I might have had to clarify some of my notation.
Feb
6
awarded  Editor
Feb
6
comment Finding the order of elements in a group?
The first part of my question IS the question I am asking. I simply do not know how to begin, and I am trying to seek help as well as provide background info on what we have learned prior to this.
Feb
6
revised Finding the order of elements in a group?
added 260 characters in body
Feb
6
comment Finding the order of elements in a group?
$Ker(f) = \{(x_1, x_2) | f(x_1) = f(x_2)\}$.
Feb
6
comment Finding the order of elements in a group?
Sorry, I was trying to define as much as I can while still keeping things terse. "For any sets $X, Y$ and any function $f: X \to Y$, there exists an equivalence relation defined on $X$, $Ker(f)$. Also, $X / Ker(f) = \{[x] | x \in X\}$ where for $x \in X$, $[x] = \{y \in X | f(y) = f(x)\}$".
Feb
6
asked Finding the order of elements in a group?
Jan
30
asked Fermat Numbers as a product
Jan
30
awarded  Informed
Jan
30
asked Explaining how $n = 2^r$ for $n$ prime
Jan
23
asked Powers of a greatest common denominator
Jan
16
accepted Is it always true that if $\gcd(a,b)=1$ then $\gcd(ab, c) = \gcd(a, c)\gcd(b, c)$?
Jan
16
accepted Proving $\gcd(a, c) = \gcd(b, c)$ for $a + b = c^2$
Jan
16
asked Proving $\gcd(a, c) = \gcd(b, c)$ for $a + b = c^2$
Jan
16
asked Is it always true that if $\gcd(a,b)=1$ then $\gcd(ab, c) = \gcd(a, c)\gcd(b, c)$?
Dec
6
accepted Orders of a symmetric group
Dec
5
comment Orders of a symmetric group
I see. It is much clearer to me now! Likewise, the total number of elements of order 5 is $\frac{5 \times 4 \times 3 \times 2 \times 1}{5} = 24$, yes? Thank you for the help!