Multivectors which commute with every vector in $ \mathbb G^n $

Question

Let $ \mathcal C $ be the set of multivectors which commute with every vector in $ \mathbb G^n $. Show that:
1) When n is even, then $ \mathcal C $ is the set of all scalars $ a $.
2) When n is odd, then $ \mathcal C $ is the set of all scalars plus pseudoscalars $ a + bI $.
Not able to make much headway at all, so any help would be appreciated.

Answer 1

I came up with this proof which is inelegant but hopefully correct:

I consider a general k-vector $ e_{i1}e_{i2}...e_{ik} $. If i can prove that there is a vector in $ G^n $ that it does not commute with, we can rule out all multi-vectors containing k-grade elements from $ C $.

Case 1: n is even
Consider a vector $ e_{i1} $. since n is even, it anti-commutes with the k-grade vector. $$ e_{i1}e_{i1}e_{i2}...e_{ik} = e_{i2}...e_{ik} = -e_{i1}e_{i2}...e_{ik}e_{i1} $$ because the number of anticommuting "swaps" required is odd. Therefore there are no k-vectors which commute with arbitrary vectors if $ k > 0 $. Which leaves only the scalars in $ C $.

Case 2: n is odd Consider the case where $ 0 < k < n $. For any k-vector $ e_{i1}e_{i2}...e_{ik} $ we can find a vector $ e_{i(k+1)} $ which is not already included in the k-vector. This vector anticommutes with the k-vector being considered: $$ e_{i1}e_{i2}...e_{ik}e_{i(k+1)} = - e_{i(k+1)}e_{i1}e_{i2}...e_{ik} $$ again because the number of "swaps" is odd. However in the cases of $ k = 0 $ and $ k = n $ the above doesn't hold and all vectors commute. This means that $ C $ consists of all $ a + bI $.