Proof that SU(n) is a group

1. Nov 6, 2015

spaghetti3451

1. The problem statement, all variables and given/known data

Show that the set of all $n \times n$ unitary matrices with unit determinant forms a group.

2. Relevant equations

3. The attempt at a solution

For two unitary matrices $U_{1}$ and $U_{2}$ with unit determinant, det($U_{1}U_{2}$) = det($U_{1}$)det($U_{2}$) = 1.

So, closure is obeyed.

Matrix multiplication is associative.

The identity matrix is unitary with unit determinant.

For a unitary matrix $U_{1}$ with unit determinant, 1 = det($U_{1}U_{1}^{-1}$) = det($(U_{1}$)det($U_{1})^{-1}$) = det($U_{1})^{-1}$).

Therefore, the inverse exists.

Am I correct?

2. Nov 6, 2015

Staff: Mentor

Your first version was better. You just had to add that 1 is unitary and unitary matrices are invertible.

3. Nov 6, 2015

spaghetti3451

How does the fact that 1 is unitary prove anything?

Same goes for the fact that the unitary matrices are invertible.

4. Nov 6, 2015

Staff: Mentor

You need to have 1 inside the group. And you showed that the inverse is unitary. What you did not show was, that the inverse exists at all. Of course it does, since det U cannot be zero. (As I said, it was almost perfect already. These 2 facts I mentioned are a little sophisticated I admit. But in a strong sense it has to be mentioned.)

5. Nov 6, 2015

spaghetti3451

Do I need to have 1 in the group because it is the identity element? Is that what you mean?

6. Nov 6, 2015

Staff: Mentor

Yes. I meant the unity matrix with 1. It all depends on your definition of SU(n). If you require SU(n) ⊂ GL(n) to be "the set of all invertible (n x n) matrices that ..." then you're done with the inverse element by showing it is unitary. If you require SU(n) ⊂ M(n) to be "the set of any (n x n) matrices that ..." then it's not clear a priori that unitary matrices are invertible. Perhaps U^-1 doesn't exits. You perfectly have done the hard part and showed that if it exists, then it has to be unitary, i.e. within the group. The existence hasn't been mentioned. I mean there are matrices that cannot be inverted.
The existence of the unitary matrix, the 1 in GL(n) is clear. Of course it is also unitary so 1 is even in SU(n). Just mention it. Otherwise you would have a 1 but not as part of the group. It has to be - an clearly is - in the group.

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