# doubt #Abelian group

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reference:Rosen

i think two more abelian groups are possible .

1 and 3 are given ,2 and 4 also exist .if i m wrong let me know ,thank you.

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Both the groups you found out are isomorphic to the second group given in Rosen.
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@Mk Utkarsh

Please explain what is isomorphic group..i have seen this for many times but haven't studied about it..

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Is group theory given in Rosen  Prateek Raghuvanshi ?  I never found it . :(. Can you plz tell me the edition exactly or the reference for reading group theory. Thanks

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Pawan Kumar 2 It's in Indian edition

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Thanks brother  Mk Utkarsh

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@Prateek,   Can you please explain the mapping for elements b and c from set A to set B in the given answer ?

Isomorphism : A group homomorphism that is bijective.

We proved that all 3 groups are homomorphic as well as isomorphic.

Note:  I rearranged the elements of table to show the similarities between all 3 tables.

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@Utkarsh , can you please explain how you created the function mapping between 2 sets. here $f$ is not defined . so how you mapped the element from one set to another. Please explain :)
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yes $f$ is not defined, i mapped the elements with same behavior in A to same behavior in B and C to show one to one correspondence in the given groups
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Sorry.. I am still not getting the mapping.  Can you please explain a little bit about mapping for elements b And c from set A to set B ?
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check the mapping and take group B's table and map the elements based of the mapping $A \rightarrow B$ you'll get the same mapping back of A

or vice versa, take A's table and interchange elements d and c you'll get the same table back.

Behavior of group B's d is same as group A's c

and

Behavior of group B's c is same as group A's d
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got it. Thank you :)
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very well explained :D

## Related questions

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Is Every Group of Order $P^{k}$ such that P is prime and K is positive integer ABELIAN
Consider the following statements: $\mathbf{S_1:}\;\;$If a group $\mathbf{(G,*)}$ is of order $\mathbf n$ and $\mathrm {a \in G}$ is such that $\mathrm {a^m=e}$ for some integer $\mathrm {m \le n}$ then $\mathbf m$ must divide $\mathbf n$. $\mathbf {S_2:}\;\;$If a ... $(3)\;\;\;\text{Niether}\; \mathrm{S_1}\;\text{nor}\;\mathrm{S_2}$