# Can a subgroup be not equal to the group?

Let A be the group of positive rational numbers under multiplication. Prove that A is isomorphic to a subgroup B such that A does not equal B.

So I know how prove isomorphisms.(define mapping, prove 1-1, prove onto, and prove operation-preserving). However, I am thrown by the fact that the subgroup does not equal A. Is it that the order is not equal? Thanks in advance for your help.

• I think you may be able to map 2 to 3 and 3 to 2. Commented Oct 18, 2015 at 23:09
• $x \mapsto x^2$. Commented Oct 18, 2015 at 23:32
• @RobArthan the mapping $x \mapsto x^2$ is not one-to-one nor onto. So it is not an isomorphism candidate that works. Commented Oct 19, 2015 at 3:44
• $x \mapsto x^2$ is one-to-one on $A$ (since the elements of $A$ are positive) and you don't want it to be onto (you want the image $B$ not to be equal to $A$). Commented Oct 19, 2015 at 8:04
• @RobArthan If it is not onto, how is it an isomorphism? Commented Oct 19, 2015 at 17:05

• That's a bit of a roundabout hint, but I see how it might help. Note that the OP's question is about $\Bbb{Q}_{+}$ under multiplication so $1$ is in any subgroup. Commented Oct 20, 2015 at 23:08