# Showing $SU(N)$ matrices commute with conjugate transpose

$$SU(N)$$ is the group of all $$N\times N$$ matrices that satisfy

$$\mathbb{U}^\dagger\mathbb{U}=1~~,\quad\text{and}\qquad \det \mathbb{U}=1~~.$$

Denoting the $$\mu$$-row and $$\nu$$-column entry in $$\mathbb{U}$$ as $$U^\mu_\nu$$, the unitarity constraint may be written as

$$\mathbb{U}^\dagger\mathbb{U}=1\quad\implies\quad \big( U^\dagger \big)^\nu_\mu U_\nu^\lambda=\delta_\mu^\nu~~.$$

I assume that the unitarity constraint is such that

$$\mathbb{U}^\dagger\mathbb{U}=1\quad\iff\quad \mathbb{U}\mathbb{U}^\dagger=1~~,$$

and I want to demonstrate this with the index algebra, and I am seeking input on whether or not there's a better way to show it. I will take the indexed expression and multiply from the right with $$U^\mu_\sigma$$, then assume the commutivity, and obtain a true expression as

\begin{align} \mathbb{U}^\dagger\mathbb{U}=1\quad\implies\qquad\qquad \big( U^\dagger \big)^\nu_\mu U_\nu^\lambda&=\delta_\mu^\nu\\ U^\mu_\sigma\big( U^\dagger \big)^\nu_\mu U_\nu^\lambda&=U^\mu_\sigma\delta_\mu^\nu\\ \left[U^\mu_\sigma\big( U^\dagger \big)^\nu_\mu \right]U_\nu^\lambda&=U^\mu_\sigma\delta_\mu^\nu\\ \text{Assume }U^\mu_\sigma\big( U^\dagger \big)^\nu_\mu=\delta^\nu_\sigma\quad\implies\qquad \qquad \qquad \delta^\nu_\sigma U_\nu^\lambda&=U^\mu_\sigma\delta_\mu^\nu\\ U^\lambda_\sigma&=U^\lambda_\sigma~~. \end{align}

I don't like it that I assumed $$U^\mu_\sigma\big( U^\dagger \big)^\nu_\mu =\delta^\nu_\sigma$$ in verifying this for myself. What is a better way for me to convince myself that

$$\mathbb{U}^\dagger\mathbb{U}=1\quad\iff\quad \mathbb{U}\mathbb{U}^\dagger=1~~?$$

When I try to use $$(\mathbb{U}^\dagger\mathbb{U})^\dagger=1^\dagger$$, the identity $$(AB)^\dagger=B^\dagger A^\dagger$$ does not let me cast the unitarity constraint in the desired form $$\mathbb{U}\mathbb{U}^\dagger=1$$.

From $$U^\dagger U = 1$$, you get $$U^\dagger = U^{-1}$$; then $$UU^\dagger = UU^{-1}= 1$$. You don't have to worry about anything; $$\det U = 1$$, this matrix is invertible.