I have some questions about group and order. I know what it is a group, but I don't really understand the concept of (order of group) and (order of element).
For example,
Thanks for you attention.
On
The order of a group is the number of elements in the set which satisfies the group axioms.
The order of an element, $x$, in a group is the minimal number of times we must multiply $x$ by itself (or add $x$ to itself, depending on the operation), to get back to the identity element.
Consider the integers modulo $5$, $\mathbb Z_5=\{0,1,2,3,4\}$. There are $5$ elements in this group. So, the order of the group is $5$. The identity of this group is $0$. Now, let's check out the element $2$:
$2+2=4$,
$2+2+2=6 \equiv 1$,
$2+2+2+2=8 \equiv 3$,
$2+2+2+2+2=10\equiv 0$.
So, $5$ is the smallest integer such that $5\cdot 2\equiv 0 \pmod 5$. Therefore, the element $2$ has order $5$.
On
It's actually pretty simple. The order of a group, also written $|G|$, is the number of elements in the group $G$. The order of an element $x$ from a group $G$ is the smallest positive integer such that $x^a = id$. I usually don't like wikipedia, but this explains it pretty well.
On
1. We can think about what $A_{5}$ actually is, that is, all even permutations of 5 symbols. If you think about it, what element in $A_{5}$ could possibly have order 60? For example if we look at something like the permutation $(1 2 3)$ which can be seen as $(13)(12)$ what is the order of this? Knowing that the order of disjoint cycles is the least common multiple of each cycle, how could we possibly get a permutation of order 60? And is this possible?
2. $|G| = 7$, then how is it possible to have elements of order either 14 or 10? There is a known fact that $\forall a \in G$ it is necessary that $ord(a)| |G|$ (Lagrange's theorem), so since 7 is prime, the only possibilities here are 1 and 7.
3. No, we can't make a homomorphism. First of all this $G$ is a cyclic group because its of prime order, and last time I checked $A_{5}$ is not even abelian. So if we have a homomorphism $\varphi: G \to A_{4}$ if we let $a,b \in G$ we'll see that $ab = ba$ (since $G$ is cyclic). Now then since $\varphi$ is a homomorphism we have $\varphi(ab) = \varphi(a)\varphi(b)$ but we also have $\varphi(ba) = \varphi(b) \varphi(a)$, this implies that $\varphi(a)\varphi(b) = \varphi(b)\varphi(a)$ but since $\varphi(a),\varphi(b)\in A_{4}$ and $A_{4}$ is not abelian then $\varphi$ cannot possibly be a homomorphism.
I hope this answers some of your questions. You may want to find or justify the facts I used for some of these answers because it's certainly the type of common knowledge you would need for a group theory midterm.
Order of a group is the number of elements in the group, i.e. the cardinality of the underlying set.
Order of an element $a$ is the order of the subgroup $\langle a \rangle$. In the finite case it's the smallest positive integer $k$ s.t. $a^k = e$.