Question: Show that $M_2(\mathbb{Q})$ is isomorphic to $\mathbb{Q} \times \mathbb{Q} \times \mathbb{Q} \times \mathbb{Q}$ where both are additive groups under the usual matrix and coordinatewise addition, respectively.

I know to show something is an isomorphism, you must first show it is a homomorphism by showing that there exists a function such that $f(a+b) = f(a)+f(b)$ and $f(ab)=f(a)f(b)$. And then to show it is isomorphic, you must show that said function is bijective, however I am uncertain of how to do it with these two groups.


An isomorphism of the additive groups only needs to respect the addition $$f(a+b) = f(a)+f(b)$$ Thus show that for the canonical Isomorphism $$f(A) = \pmatrix{A_{11}\\A_{21}\\A_{12}\\A_{22}}$$ $$f(A+_{M_2(\mathbb Q)}B) = f(A) +_{\mathbb Q^4} f(B)$$ With both additions element-wise.


As others have pointed out, you don't need to worry about multiplication. Since the question only asks about a group isomorphism, there is only one operation to concern yourself with.

As is, if you use the (reasonably obvious) multiplication operation on $\mathbb{Q}^{\oplus 4}$---that is, the one given by $$ (a,b,c,d) \cdot_{\mathbb{Q}^{\oplus 4}} (e,f,g,h) = (ae, bf, cg, dh) $$ then it is not a homomorphism, which is probably worthwhile to check. However, this isn't relevant, as the question is only about the category of groups, and not of rings.

  • $\begingroup$ I'm not sure why you say it isn't a ring; it is a ring, but it isn't a field. $\endgroup$ – Simon Rose Dec 1 '14 at 10:15
  • $\begingroup$ Yeah sorry mixed the terms. I should drink some coffee. $\endgroup$ – AlexR Dec 1 '14 at 12:09
  • 1
    $\begingroup$ You should always drink some coffee! $\endgroup$ – Simon Rose Dec 1 '14 at 12:29

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