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I am not sure if this question has been asked before, but my search did not return me any answers. While reading several online notes that attempt to give an intuitive understanding of group homomorphisms, I have found people drawing similarities between group homomorphism and linear transformation. However, I still do not fully understand how these two ideas are related. Could someone briefly explain how these two concepts are related? If these two are useful analogies of each other, what would group isomorphism be analogous to in linear algebra?

Four Seasons
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    Possibly helpful: https://math.stackexchange.com/questions/2039702/what-is-an-homomorphism-isomorphism-saying/2039715#2039715 – Ethan Bolker Mar 24 '19 at 14:48
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    A linear transform between vector spaces is in particular a group homomorphism between the additive groups of the two vector spaces. In addition, it takes care of scalar multiplication. – amsmath Mar 24 '19 at 15:06

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By definition, the element in the vector space equipped with the vector space addition is a group as it satisfies all the axioms of a group addition. A homomorphism between two groups $G$ and $H$ is an operation preserving mapping $f :G \rightarrow H$. Namely, For any element $v, u$ in the domain group, $f(v+u)=f(u)+f(v)$. So in other words, it doesn't matter if we first add the two elements $u,v$ using the addition of the domain group and then map the result to the image group as $f(u+v)$ or map them first to the image group using the homomorphism as $f(u),f(v)$ and then add them using the addition operation on the image group as $f(u)+f(v)$. You will get the same answer in both ways.

Well, Vector spaces on the other hands are groups, but they have additional elements/properties. As not only they have an addition operation, but they are also equipped with a field/scaler usually the real numbers or the complex numbers, together with a field/scalar multiplication operation, which can be thought of a map $s: V\times R \rightarrow G$. Namely, $s$ takes as input a scalar $s1$ and a vector $v$ and produces another vector $s(v,s1)$. And a linear map not only has to preserve the addition of the vector spaces like when a homomorphism preserve the addition operation defined on the two groups, but it also has to preserve the multiplication operation in the same sense above. Like $f(s(v,s1))=s(f(v),s1)$.

user614287
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If you "forget" that multiplication by scalar exists on a vector space, you will still have a group with the addition operation. In this way, vector spaces are a special kind of groups, equipped with one more algebraic structure given by multiplication by scalar.

When you think of linear transformations $T:U \to X$, they satisfy: \begin{align} T(u+v) &= T(u) + T(v) \\ T(\alpha u) &= \alpha T(u) \end{align}

That means that $T$ preserves the algebraic structure, that is, it preserves the operations. It doesn't matter if you sum the vectors or multiply them by scalars before or after applying $T$, the result is the same.

Now group homomorphisms $\phi:G \to M$ satisfies $\phi(gh) = \phi(g)\phi(h)$. If the group operation is addition (which is specially denoted with $+$), you simply get $\phi(g+h) = \phi(g)+\phi(h)$, recovering the sum of vectors under $T$ in vector spaces.

So, as one might expect, linear transformations is a special kind of group homomorphism because they also preserve the previous group structure (AKA addition) that is already present in vector spaces. But they also preserve something more, the multiplication by scalar. So just like vector spaces are "glorified" groups, linear transfomations are also "glorified" group homomorphisms.

AspiringMathematician
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