Yes, #2, although I would formulate it in terms of group theory: ##\frac{\pi}{2} \in \langle \mathbb{R}_{2\pi}, +_{2\pi}\rangle ## is of order ##4## and ##\langle \mathbb{R},+\rangle ## has no elements of finite order. Can you show, that this implies ##\langle \mathbb{R}_{2\pi}, +_{2\pi}\rangle \ncong \langle \mathbb{R},+\rangle\; ##?Mr Davis 97 said:So, for example, if I wanted to show that the latter is not isomorphic to the former, could I just note that the equation ##z +_{2 \pi} z +_{2 \pi} z +_{2 \pi} z =0## has four solutions ##\{0,\frac{\pi}{2}, \pi, \frac{3 \pi}{2} \}##, while the equation ##x+x+x+x=0## has only one solution, ##\{ 0 \}##? And argue that if they were isomorphic then these equations should have the same number of solutions? Is this an example of an invariant you were describing?
Yes, but the "work" to be done is in the first sentence. Why do they preserve the order?Mr Davis 97 said:Well, isomorphisms preserve order, right? So if there were an isomorphism, that would imply that ##\langle \mathbb{R},+\rangle## has at least one element of order 4. However, it has no elements of finite order, which means that there cannot exist an isomorphism.
Bashyboy said:How are you defining ##\mathbb{R}_{2 \pi}## and ##+_{2 \pi}##?
Addition of angles modulo ##2\pi##, I think, ##U(1)\;.##Bashyboy said:I hope this isn't too much of an imposition, but I am still interested in an answer to this question.
To determine if two groups are isomorphic, you must show that there exists a bijective homomorphism between them. This means that there is a one-to-one correspondence between the elements of the two groups, and the operation of the first group can be mapped to the operation of the second group in a way that preserves the group structure.
Yes, two groups can have the same number of elements but still be non-isomorphic. This is because the group structure and operation may be different, even though the number of elements is the same. For example, the cyclic group of order 4 and the Klein four-group both have 4 elements, but they are not isomorphic.
There are several methods for proving two groups are not isomorphic. Some common ones include showing that the orders of the groups are different, demonstrating that the groups have different subgroups or different number of subgroups, and showing that the groups have different properties (e.g. commutative vs non-commutative).
No, isomorphic groups must have the same Cayley tables. The Cayley table represents the group operation and structure, so if two groups are isomorphic, they must have the same Cayley table. This is why the Cayley table is a useful tool for determining whether two groups are isomorphic.
Yes, it is possible for two non-isomorphic groups to have the same order. This is because the order of a group only tells us the number of elements in the group, but it does not provide information about the group structure or operation. For example, the symmetric group S3 and the dihedral group D3 both have 6 elements, but they are not isomorphic.