Isomorphic groups

1. Jul 4, 2014

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?

Last edited: Jul 5, 2014
2. Jul 4, 2014

micromass

Staff Emeritus
I really don't understand your first sentence.

3. Jul 5, 2014

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?

4. Jul 5, 2014

micromass

Staff Emeritus
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.

5. Jul 5, 2014

LagrangeEuler

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

6. Jul 5, 2014

NeoAkaTheOne

$\mathbb{Z}_p$ (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.

7. Jul 5, 2014

micromass

Staff Emeritus
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 $2$.

8. Jul 5, 2014

LagrangeEuler

Yes. Exactly!

9. Jul 5, 2014

LagrangeEuler

Tnx. For this definitions. I will read that.

10. Jul 5, 2014

micromass

Staff Emeritus
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 [Broken])
http://en.wikipedia.org/wiki/Modular_group

Last edited by a moderator: May 6, 2017
11. Jul 5, 2014

NeoAkaTheOne

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.

12. Jul 5, 2014

micromass

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

13. Jul 5, 2014

NeoAkaTheOne

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 $8$?

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