How can there only be two possible four-element groups?

[Azure Ace]

How can you prove that there can only be 2 possible four-element group?

 

[fresh_42]

E.g. with the help of Lagrange's theorem. But try and error with a group table will do with such a small example.

 

[Azure Ace]

I was trying to prove this and thought that the proof would be as simple as determining the number of possible element orders. Like for group order 4, the possible element orders are 2 and 4. Then, the possible groups would either have an element with order 2 or 4. Since there are only 2 possible element orders, there can only be 2 four-element groups: one whose generator element has order 2 and the other with order 4. Is this a valid proof or am I jumping to conclusions? Can there be only one element order in a group?

 

[fresh_42]

This is basically correct. An element of order $4$ leads to $\mathbb{Z}_4$. So there is only the possibility left, where we have only elements of order $2$. Basically this can lead to the group $\mathbb{Z}_2^2$ and groups $\mathbb{Z}_2 \rtimes_\varphi \mathbb{Z}_2$. Now we need an argument, why $\varphi = 1$ is the only possibility, i.e. why all semidirect products are already direct.

To your second question: What do you mean by "one element order"? If you mean, whether all group elements have to be of the same order, then the answer is no. E.g. $\mathbf{Sym}(3) = \langle (1),(12),(123) \rangle$ has elements of order $2$ and $3$. Also $1 \in \langle \mathbb{Z}_4 , + \rangle$ is of order $4$ whereas $2 \in \langle \mathbb{Z}_4 , + \rangle$ is of order $2$

 

[Azure Ace]

This is basically correct. An element of order $4$ leads to $\mathbb{Z}_4$. So there is only the possibility left, where we have only elements of order $2$. Basically this can lead to the group $\mathbb{Z}_2^2$ and groups $\mathbb{Z}_2 \rtimes_\varphi \mathbb{Z}_2$. Now we need an argument, why $\varphi = 1$ is the only possibility, i.e. why all semidirect products are already direct.

To your second question: What do you mean by "one element order"? If you mean, whether all group elements have to be of the same order, then the answer is no. E.g. $\mathbf{Sym}(3) = \langle (1),(12),(123) \rangle$ has elements of order $2$ and $3$. Also $1 \in \langle \mathbb{Z}_4 , + \rangle$ is of order $4$ whereas $2 \in \langle \mathbb{Z}_4 , + \rangle$ is of order $2$

It is my first time taking a group theory course and so far, we have only dealt with group properties, group tables, $\mathbb{Z}_n$, and the order or a group and element. So I don't understand yet some of the parts you are talking about but I still appreciate your answers and will consider them in the future when we get to that point.

 

[fresh_42]

I guess you don't mean the order of elements part. In this case, $(n_1,n_2,n_3,\ldots ,n_k)$ denotes a permutation: $n_i \mapsto n_{i+1}$ and $n_k \mapsto n_1$. These build the so called symmetric groups, in my example of "order" three. "Order" is a misleading term here, as it means "all permutations of maximal three elements", which is neither the group order $|\mathbf{Sym}(3)|=6$ nor the order of elements $\operatorname{ord}(12)=2\, , \,\operatorname{ord}(123)=3$

The other point has been the semidirect products. Let me try another argument here. We have a group $G$ with $|G|=4$ and only elements $a_i \in G$ with $a_i^2=1$. This follows from Lagrange, since an element of order $4$ automatically leads to the cyclic group of order $4$, so $G=\mathbb{Z}_4$. Therefore we have $G=\{1,a_1,a_2,a_3\,\vert \,a_i^2=1\}$. Now use the group properties to figure out the possible values for $a_1\cdot a_2\; , \;a_2 \cdot a_3\; , \; a_1 \cdot a_3$ and why the group has to be Abelian.

 

[lpetrich]

If you wish to do it with multiplication tables, there are two properties of groups that will make your life easier, so that you do not have to experiment as much with tables.

1. There is only one identity element
2. Every row in the table is a permutation of the elements, with no elements stationary from row to row. Likewise for columns, though a row and a column may have the same permutation.

These two properties can easily be proved from the definition of a group. For the first one, if there are two identities, what happens when you multiply them? For the second one, if a.b = a.c, then how are b and c related? Likewise for b.a = c.a.

For a 4-element group, then Lagrange's theorem means that every possible non-identity element order must be either 2 or 4, as others here have pointed out. If it is 4, then we are done: Z4. If it is 2, then it is a bit trickier. For elements e, a, b, c, the only possible value of a.b is c, and likewise for b.a. That means that the group is commutative. Thus, the group is Z2 * Z2, with the Z2 generators being a and b, a and c, or b and c.

Here are some theorems that one can easily prove with Lagrange's theorem:

1. Every prime-order group is cyclic.
2. No element of a nonabelian group has an order equal to the group's order. The converse is not necessarily true, and it is easy to find counterexamples to it.