Isomorphic Groups: Z2 X Z3 & G = (1,2,4,8,10)

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Discussion Overview

The discussion revolves around the isomorphism between the group Z2 x Z3 and another group G, initially defined as (1,2,4,8,10). Participants explore the properties of these groups, including their elements and structure, in the context of abstract algebra.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that Z2 x Z3 cannot be isomorphic to G due to differing numbers of elements (6 vs. 5).
  • Others propose that the question is incorrect and suggest an alternative group G = (1,2,4,5,7,8) with respect to multiplication modulo 9.
  • One participant emphasizes the importance of comparing elements of order two in both groups to determine isomorphism.
  • Another participant claims that since both groups have 6 elements, they must be isomorphic and begins to outline a proof involving their elements.
  • There is a suggestion to check if G is cyclic by verifying if [2]_9 serves as a generator.
  • A later post indicates a misunderstanding regarding the notation for G, which led to a clarification about the nature of the groups involved.
  • One participant mentions using Cayley's theorem to find a subgroup of S5 that is isomorphic to Z2 x Z2.

Areas of Agreement / Disagreement

Participants express disagreement regarding the initial formulation of the problem and the isomorphism between the groups. Multiple competing views remain about the correct definition of G and the conditions for isomorphism.

Contextual Notes

There are unresolved assumptions regarding the definitions of the groups and the nature of their elements, particularly concerning the order of elements and the structure of the groups involved.

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hi

Show that Z2 X Z3 IS ISOMORPHIC TO THE GROUP G = (1,2,4,8,10)
 
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Re: abstract algebra

rss said:
hi

Show that Z2 X Z3 IS ISOMORPHIC TO THE GROUP G = (1,2,4,8,10)

Hi rss! Welcome to MHB! (Smile)

Which group does G represent?

Either way, since Z2 x Z3 contains 6 elements, and G contains 5 elements, they cannot be isomorphic.
 
Re: abstract algebra

I like Serena said:
Hi rss! Welcome to MHB! (Smile)

Which group does G represent?

Either way, since Z2 x Z3 contains 6 elements, and G contains 5 elements, they cannot be isomorphic.

hi
this question is incorrect.
the correct question is
show that the group Z2xZ3 is isomorphic to the group G=(1,2,4,5,7,8) with respect to multiplication modulo 9
 
Re: abstract algebra

rss said:
hi
this question is incorrect.
the correct question is
show that the group Z2xZ3 is isomorphic to the group G=(1,2,4,5,7,8) with respect to multiplication modulo 9

Hi rss,

Consider the elements of order two in $\Bbb Z_2 \times \Bbb Z_3$ and $G$. If the groups don't have the same number of elements of order two, then they are not isomorphic.
 
Re: abstract algebra

Euge said:
Hi rss,

Consider the elements of order two in $\Bbb Z_2 \times \Bbb Z_3$ and $G$. If the groups don't have the same number of elements of order two, then they are not isomorphic.

there are a total of 6 elements in each group so they must be isomorphic. i just have to prove them so
Z2xZ3= (0,0),(0,1),(0,2),(1,0),(1,1),(1,2)
and G=(1,2,4,5,7,8)...0N MULTIPLICATION with respect to modulo 9 we get
x 1 2 4 5 7 8
1 1 2 4 5 7 8
2 2 4 8 1 5 7
4 4 8 7 2 1 5
5 5 1 2 7 8 4
7 7 5 1 8 4 2
8 8 7 5 4 2 1

how do i proceed from here
 
Re: abstract algebra

rss said:
there are a total of 6 elements in each group so they must be isomorphic.

Sorry I misread your post, your notation for $G$ looked like cycle notation. The groups are isomorphic because they they are abelian groups of order $6$ (the symmetric group on $3$ letters is a nonabelian group, which cannot be isomorphic to an abelian group). Note that $\Bbb Z_2 \times \Bbb Z_3$ is cyclic, generated by $([1]_2, [1]_3)$. So it suffices to show that $G$ is cyclic. Check that $[2]_9$ is a generator of $G$.

It turns out in fact that every abelian group of order $6$ is a cyclic group.
 
Re: abstract algebra

rss said:
there are a total of 6 elements in each group so they must be isomorphic. i just have to prove them so
Z2xZ3= (0,0),(0,1),(0,2),(1,0),(1,1),(1,2)
and G=(1,2,4,5,7,8)...0N MULTIPLICATION with respect to modulo 9 we get
x 1 2 4 5 7 8
1 1 2 4 5 7 8
2 2 4 8 1 5 7
4 4 8 7 2 1 5
5 5 1 2 7 8 4
7 7 5 1 8 4 2
8 8 7 5 4 2 1

how do i proceed from here

using Cayleys theorem find a subgroup of S5 to which Z2xZ2 is isomorphic
 
A new question should be put in a new post. Also, show effort on working the problem to receive better assistance.
 

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