I need to come up with two groups that are bijective, but not isomorphic. So I thought about the map $L_g: G \rightarrow G'$. $L_g(x) = gx$.
$L_g: x \rightarrow gx$ Correct
$L_{g^-1}: gx \rightarrow x$ Doesn't work.
I am a little confused why the group is not isomorphic. Any help is much appreciated.
On
You can take two groups with the same number of elements and different structural features. This is a good strategy for showing any two groups are not isomorphic. One good class of differentiating structural features are those that arise from the group's binary operation. One that might come to mind is commutativity. Certainly, if one group is abelian, and the other nonabelian, no homomorphism can be found between the two.
So, for one of our groups, let's take $S_3$, the symmetry group with $3! = 6$ elements. This group is nonabelian (if you don't know this, it's easy to exhibit via the Cayley diagram of $S_3$). Then take $\langle \mathbb{Z}_6, + \rangle$, the integers modulo six equipped with integer addition. We know that $\mathbb{Z}_6$ is abelian, as integer addition is commutative. But $S_3$, as I mentioned, is nonabelian. However, they both have six elements, so a bijection exists between their sets. But no homomorphism can ever exist, so they are not isomorphic.
Take the set of numbers $G=\{1,3,5,7\}$ with binary operation being multiplication modulo 8. Check that this is a group. Now consider $U=\{1,-1, +i, -i\}$; this is also a group, but now for the operation of multiplication of complex numbers.
As both have 4 elements they are bijective. Any bijection one can construct between these two groups will never be a homomorphism. (one is cyclic other is not).