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

[rss1]

hi

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

 

[I like Serena]

Re: abstract algebra

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.

 

[rss1]

Re: abstract algebra

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

 

[Euge]

Re: abstract algebra

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.

 

[rss1]

Re: abstract algebra

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

 

[Euge]

Re: abstract algebra

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.

 

[rss1]

Re: abstract algebra

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

 

[Euge]

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