GloVe vector representation homomorphism question
If you're asking if the group homomorphism makes the the process symmetric then no it doesn't directly.  However, they use the fact that they require a group homomorphism to show that $w_{i}^{T} \tilde{w}_k = log(P_{ik})=log(X_{ik}) - log(X_{i})$  This nearly gives us symmetry.  Finally by adding $\tilde{b}_{k}$ into the equation you restore symmetry.  

So in short $w_{i}^{T} \tilde{w}_k + b_{i} + \tilde{b}_{k} = log(X_{ik})$ is what ensures symmetry, and the group homomorphism is a tool to get there.

Update:

Some more details
Essentially, what we want is the ability to peform a label switch.  Group homomorphism helps with this process because it perseves a mapping between the $(R, +)$ and $(R, x)$.  

$F((w_{i}^{T} - w_{j}^{T})w_{k}^{'})=F(w_{i}^{T}w_{k}^{'}+( - w_{j}^{T}w_{k}^{'})) = F(w_{i}^{T}w_{k}^{'}) \times F(-w_{j}^{T}w_{k}^{'} )= F(w_{i}^{T}) \times F(w_{j}^{T}w_{k}^{'})^{-1} = \frac{F(w_{i}^{T}w_{k}^{'})}{F(w_{j}^{T}w_{k}^{'})}$

The group homomorphism here allows for that to occur.   Therefore we can see that by setting $F(w_{i}^{T}w_{k}^{i}) = \frac{X_{ik}}{X_{i}}$

Now finally we can say that $w_{i}^{T} {w}_k^{'} = log(P_{ik})=log(X_{ik}) - log(X_{i}).$

So as far as your comment, it is the most sensible chocie for their method and of which they buld the core mathematicals to GloVE.  Changing it, I imagine wouldn't be a trivial thing.  I imagine if you did, much of what is derived, including the loss function would change.  But with that said, I imagine there are otherwise to achieve label switching.