Show that Z_12^* and Z_8^* are isomorphic groups

  • Thread starter Thread starter Mr Davis 97
  • Start date Start date
  • Tags Tags
    Groups
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
1 reply · 2K views
Mr Davis 97
Messages
1,461
Reaction score
44

Homework Statement


Show that ##\mathbb{Z}_8^*## and ##\mathbb{Z}_12^*## are isomorphic, where ##\mathbb{Z}_n^* = \{x \in \mathbb{Z} ~|~ \exists a \in \mathbb{Z}_n(ax \equiv 1~(mod~n)) \}##, and the group operation is regular multiplication.

Homework Equations

The Attempt at a Solution


We can see that ##\mathbb{Z}_8^* = \{1,3,5,7 \}## and ##\mathbb{Z}_8^* = \{1,5,7,11 \}##

The only possible isomorphism I can think if is a function that maps from the former to the latter such that 1 goes to 1, 3 to 5, 5 to 7, and 7 to 11. The function is obviously injective and surjective. Is the only way to show that this satisfies the homomorphism property to show that it is satisfied for each combination from the domain? This would seem to be a tedious process.
 
Physics news on Phys.org
The quick way is to note that up to isomorphism there are two abelian groups of order 4, the cyclic group [itex]C_4[/itex] (which contains an element of order 4) and the klein group [itex]C_2 \times C_2[/itex] (which does not contain an element of order 4).