Proving $SU(n-1) $is a subalgebra of $SU(n)$ algebra. I am a physics student trying to learn as much group theory as possible as it is pertinent to my studies in High Energy Physics. I can see that $SU(2)$ is contained in $SU(3)$ but is it true that $SU(n-1)$ is a subalgebra of $SU(n)$ algebra? If so, how would I be able to show it?
 A: There's an easy way to see this. We know that an $n\times n$ matrix is in $SU(n)$ if and only if $A^{\dagger}A=I_n$, where $I_n$ denotes the $n\times n$ identity matrix. We can embed $SU(n-1)$ into $SU(n)$ as follows. Take an $(n-1)\times (n-1)$ matrix in $SU(n-1)$, denoted by $B$. Then send it to $B'$ defined by 
$$ B'=
\begin{bmatrix}
B&0\\
0&1
\end{bmatrix}$$
where $B$ is to be understood as an $(n-1)\times (n-1)$ square block. This defines an injective group homomorphism $SU(n-1)\to SU(n)$. Indeed, note that 
$$ (B')^{\dagger}=
\begin{bmatrix}
B^\dagger&0\\
0&1
\end{bmatrix}=
\begin{bmatrix}
B^{-1}&0\\
0&1
\end{bmatrix}=B'^{-1}$$
by the assumption on $B$. So, this lets us identify $SU(n-1)$ with a subgroup of $SU(n)$. If you know what a manifold is, then it isn't too hard to show that $SU(n-1)$ is a submanifold of $SU(n)$. Noting, then, that $\mathfrak{su}(n)=T_{I_n} SU(n)$, it follows that $\mathfrak{su}(n-1)\subseteq \mathfrak{su}(n)$, since in general for a submanifold $N\subseteq M$, and a point $p\in M$, we have $T_pN\subseteq T_pM$ as a linear subspace.
If you don't want to deal with manifolds, then you can show this latter fact using the method in Anz3141's answer.
A: I think you mean to ask if $SU(n-1)$ is a subgroup of $SU(n)$.
Yes. Since we can equivalently work with the corresponding Lie algebras, we can show that $\mathfrak{su}(n-1)$ is a subalgebra of $\mathfrak{su}(n)$. First, $\mathfrak{su}(n)$ can be faithfully represented by the algebra of n-by-n traceless skew-Hermitian matrices. Now, consider the subset of these matrices for which the last row and column are set to $0$. The upper-left (n-1)-by-(n-1) block is itself a traceless skew-Hermitian matrix. You can check easily that this subset is a subalgebra.
Since the representation is faithful, $\mathfrak{su}(n-1)\subset\mathfrak{su}(n)$ and hence $SU(n-1)\subset SU(n)$. The inclusion symbols represent "Lie subalgebra" and "Lie subgroup", respectively.
