- #1

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

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- Thread starter Azure Ace
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- #1

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

- #2

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- #3

- #4

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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##

- #5

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.

- #6

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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.

- #7

lpetrich

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- There is only one identity element
- 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.

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:

- Every prime-order group is cyclic.
- 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.

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