# First Isomorphism Theorem Example

In this post, I want to talk about an application of the First Isomorphism Theorem to linear transformations between different dimensions. We’ll see why it’s kind of surprising why this works at all, but also why it makes sense that it works.

Let’s start by stating the version of the First Isomorphism Theorem that we’ll be talking about.

First Isomorphism Theorem for Groups: Let $\phi: G\to H$ be a surjective homomorphism. Then $H$ is isomorphic to $G/ker(\phi)$.

Note that since $\phi$ is surjective (onto), we can just say $H$ instead of $im(\phi)$, since every element of $H$ gets mapped to by $\phi$.

Example: Consider the transformation $y =\begin{pmatrix}1&0&0\\0&1&0\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\begin{pmatrix}x_1\\x_2\end{pmatrix}$.

To apply the First Isomorphism Theorem to this example, we let $\phi$ be this linear transformation, where the group composition law is vector addition, satisfying $\phi(a+b)=\phi(a)+\phi(b)$.

The kernel of this linear transformation is a vector of the form $\begin{pmatrix}0\\0\\k\end{pmatrix},$ for any $k\in\mathbb{R}$.

Since this transformation goes from $\mathbb{R}^3$ to $\mathbb{R}^2$, to examine $G/ker(\phi)$, we need to figure out what $\mathbb{R}^3/ \begin{pmatrix}0\\0\\k\end{pmatrix}$ looks like. This is like asking, “what does everything in $\mathbb{R}^3$ that has a zero in the third position look like?”

This means that we can write $G/ker(\phi)$ as span{$\begin{pmatrix}1\\0\\0\end{pmatrix},\begin{pmatrix}0\\1\\0\end{pmatrix}$}.

The First Isomorphism Theorem says that this should be isomorphic to $\mathbb{R}^2$. This makes a lot of sense, because the vectors in the set above are the $x$ and $y$ unit vectors (so they definitely span $\mathbb{R}^2$). Even though the result of taking the quotient group above lives in $\mathbb{R}^3$, the only thing that this theorem cares about is the fact that the resulting set after taking the quotient is a two-dimensional plane.