Isomorphic Groups: Proving Theorem True?

[LagrangeEuler]

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?

 

[micromass]

I really don't understand your first sentence.

 

[LagrangeEuler]

Sorry Sir. I try to say. If i had to groups with same number of elements $|G_1|=|G_2|$ and groups $(G_1,\cdot)$ and $(G_2,*)$ has the same number of element with the same order are then the groups are isomorphic? Is there such theorem?

 

[micromass]

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]

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.

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]

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
$(\mathbb{Z}_p\times\mathbb{Z}_p)\rtimes\mathbb{Z}_p$?

$\mathbb{Z}_p$ (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]

$\mathbb{Z}_p$ (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.

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 $2$.

 

[LagrangeEuler]

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 $2$.

Yes. Exactly!

 

[LagrangeEuler]

$\mathbb{Z}_p$ (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.

Tnx. For this definitions. I will read that.

 

[micromass]

Yes. Exactly!

Well, it's not true. The simplest possible example involve the group $\mathbb{Z}_2\times\mathbb{Z}_8$ and the modular group. These are groups of order 16.

http://en.wikipedia.org/wiki/Cycle_graph_(algebra )
http://en.wikipedia.org/wiki/Modular_group

 

[NeoAkaTheOne]

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 $2$.

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]

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.

I think $\mathbb{Z}/8\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ has an element of order 8 though.

 

[NeoAkaTheOne]

I think $\mathbb{Z}/8\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ has an element of order 8 though.

You're correct. Small oversight, but I think I meant $Q\times \mathbb{Z}/2\mathbb{Z}$ with $Q$ being the quaternion group. Surely that doesn't have an element of order $8$?