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LagrangeEuler
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Two groups are isomorphic if they has same number of elements and if they has same number of elements of same order? Is it true? Where can I find the prove of this theorem?
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Tnx a lot and what are that group. What is ##\mathbb{Z}_p##? And what is difference betweenmicromass said:No, there is no such theorem because it is false. I think the two groups ##\mathbb{Z}_p\times\mathbb{Z}_p\times\mathbb{Z}_p## and ##(\mathbb{Z}_p\times\mathbb{Z}_p)\rtimes\mathbb{Z}_p## constitute a counterexample for ##p## prime.
LagrangeEuler said:Tnx a lot and what are that group. What is ##\mathbb{Z}_p##? And what is difference between
##\mathbb{Z}_p\times\mathbb{Z}_p\times\mathbb{Z}_p##
and
##(\mathbb{Z}_p\times\mathbb{Z}_p)\rtimes\mathbb{Z}_p##?
NeoAkaTheOne said:##\mathbb{Z}_p## (also denoted as the quotient group ##\mathbb{Z}/p\mathbb{Z})## is the set of residue classes modulo ##p##. So for example, ##\mathbb{Z}_5=\{0,1,2,3,4\}##. In the above example, ##\rtimes## denotes the semi direct product, as opposed to ##\times## which is the direct product. See http://en.wikipedia.org/wiki/Direct_product and http://en.wikipedia.org/wiki/Semidirect_product for good explanations of these.
The answer above is not likely to be of help to you, since your question implies that you are a novice in group theory. A more fitting answer for a counterexample is to consider ##\mathbb{Z}/6\mathbb{Z}## and ##S_3##, the group of permutations of ##3## elements. They both have order ##6##, and they certainly are not isomorphic.
micromass said:I think the OP wants two groups ##G## and ##G^\prime## that have the same order, and such that ##G## and ##G^\prime## have the same number of elements with a given order. For example, ##G## and ##G^\prime## both have the same number of elements of order ##2##.
NeoAkaTheOne said:##\mathbb{Z}_p## (also denoted as the quotient group ##\mathbb{Z}/p\mathbb{Z})## is the set of residue classes modulo ##p##. So for example, ##\mathbb{Z}_5=\{0,1,2,3,4\}##. In the above example, ##\rtimes## denotes the semi direct product, as opposed to ##\times## which is the direct product. See http://en.wikipedia.org/wiki/Direct_product and http://en.wikipedia.org/wiki/Semidirect_product for good explanations of these.
The answer above is not likely to be of help to you, since your question implies that you are a novice in group theory. A more fitting answer for a counterexample is to consider ##\mathbb{Z}/6\mathbb{Z}## and ##S_3##, the group of permutations of ##3## elements. They both have order ##6##, and they certainly are not isomorphic.
LagrangeEuler said:Yes. Exactly!
micromass said:I think the OP wants two groups ##G## and ##G^\prime## that have the same order, and such that ##G## and ##G^\prime## have the same number of elements with a given order. For example, ##G## and ##G^\prime## both have the same number of elements of order ##2##.
NeoAkaTheOne said:Ah, the question was not very eloquently phrased.
To answer the question, mircomass is correct in saying that the smallest example comes with order ##16##. Consider ##\mathbb{Z}/8\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}\not\simeq\mathbb{Z}/4\mathbb{Z}\times\mathbb{Z}/4\mathbb{Z}\not\simeq H## where ##H=<a,b>## such that ##|a|=|b|=4##. It's quite clear that these three groups have ##12## elements of order ##4## and ##3## elements of order ##2## (and obviously the identity being the sixteenth). Is this perfect?
More complicated examples can be arrived at by the notion of ##p##-groups, which I imagine OP has not studied yet.
micromass said:I think ##\mathbb{Z}/8\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}## has an element of order 8 though.
An isomorphic group is a mathematical concept that refers to two groups that have the same structure, even if the elements and operations may be different. In simpler terms, it means that the groups are essentially the same, just expressed differently.
To prove that two groups are isomorphic, you must show that there is a one-to-one correspondence between the elements of the two groups, and that the group operations are preserved. This can be done by demonstrating a bijective function between the two groups that preserves the group structure.
Proving the isomorphic theorem is important because it allows us to show that seemingly different groups are actually the same. This can help simplify calculations and provide insights into the properties of the groups.
No, not all groups are isomorphic. In order for two groups to be isomorphic, they must have the same structure. This means that the order of the elements, the group operations, and the identity element must all be the same.
Yes, the isomorphic theorem can be applied to more than two groups. In fact, it can be used to compare and classify multiple groups at once, by showing that they are all isomorphic to a certain group. This can help organize and understand the properties of various groups.