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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,

  1. $A_5$ (alternating group) has an element of order 60? Why?
  2. A group G of order 7 (I don't know what this mean) could have a element of order 14? And element of order 10? Why?
  3. Could we make an Group homomorphism between $A_5$ and $G$? Why?

Thanks for you attention.

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  • $\begingroup$ Where did you read the second statement? That is false. What does degree mean in the first statement? $\endgroup$ – Sahiba Arora Jan 21 '18 at 18:12
  • $\begingroup$ One moment, I edit. $\endgroup$ – mathandtic Jan 21 '18 at 18:13
  • $\begingroup$ No element in $A_5$ has order $60$. The order of a finite group is just the number of elements in the group. The order of an element $g$ in a finite group is the least natural number $n$ such that $g^n=e$. $\endgroup$ – lulu Jan 21 '18 at 18:15
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    $\begingroup$ Because $A_5$ isn't cyclic, it isn't even abelian. $\endgroup$ – lulu Jan 21 '18 at 18:17
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    $\begingroup$ Why start with such complicated groups? If you are puzzled by the basic definitions, start with simpler groups. Take $S_3$, for example. The non-abelian group of order $6$. Write it out carefully and compute the order of each element. $\endgroup$ – lulu Jan 21 '18 at 18:19
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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.

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    $\begingroup$ Yes, also remember that the minimality requirement of the order implies that that order of an element is less than or equal to the order of the group. This comes from the fact that the order of every element of the group must also divide the order of the group. So clearly you can't gave an order $k$ of the element $a$ such that $ k > |G|$. $\endgroup$ – Boots Jan 21 '18 at 18:43
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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. $a^k = e$.

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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.

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  • $\begingroup$ For example, if order of $G$ is 20, does it exist a element of order 5? I think yes due to Lagrange's Theorem, but I'm not sure. $\endgroup$ – mathandtic Jan 21 '18 at 19:02
  • $\begingroup$ Remember that the converse of Lagrange's Theorem is not always true. If $|G| = 20$ it does have an element of order 5, but this is by Cauchy's Theorem which states that for every prime $p$ dividing $|G|$, $G$ contains an element of order $p$. Lagrange's Theorem states that the element of a group has to divide the order of the group, but if a number divides the order of the group, it is not necessary to have an element of that order. $\endgroup$ – Boots Jan 21 '18 at 19:06
  • $\begingroup$ I have a group $G$, which order is $a$ (number of elements), if I could find a prime number $p$ that divides the order of $G$ then I could find a element $a$ with order $p$, isn't it? $\endgroup$ – mathandtic Jan 21 '18 at 19:10
  • $\begingroup$ Try not to use $a$ to denote two different things, but yes this is the statement of Cauchy's Theorem. $\endgroup$ – Boots Jan 21 '18 at 19:23
  • $\begingroup$ If I have a group $G$, which order 62, what Lagrange's Theorem said about it? I know that Cauchy said that exists a element $b$ of order 31. $\endgroup$ – mathandtic Jan 21 '18 at 19:28
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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 $5$.

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  • $\begingroup$ The order of 3 is 1? (because $3+3=6 = 1$) $\endgroup$ – mathandtic Jan 21 '18 at 20:43
  • $\begingroup$ No, this is addition modulo 5, where the identity is 0. So you would need $3 + 3 + 3 + 3 + 3 = 15 \equiv 0$ which would mean the order of 3 is 5. $\endgroup$ – Boots Jan 22 '18 at 12:32

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