I am having some difficulty with this problem. We are asked to show that there does not exist any monomorphism to GL(n, $\mathbb{C}$). The group in the domain is defined as follows:
Let G = $(P(\mathbb{R}), \delta)$ where $\delta$ is the symmetric difference operator, and $P(\mathbb{R})$ is the power set of the reals.
I noted that, for all $A \subseteq P(\mathbb{R})$, we have the following:
A$\delta$A = $\emptyset$
A$\delta \emptyset$ = A
A$\delta\mathbb{R}$ = A$^C$
I am not really sure how to use these facts, but the strategy I initially considered was proof by contradiction. I can't seem to find any way to show that any properties of homomorphisms or injectivity are violated, though.
I briefly thought of using a cardinality argument, but I am not sure how to proceed with that either.
Any help would be greatly appreciated. Even just a hint would be awesome. Thanks!
$G$ has cardinality $\beth_2=2^{2^{\aleph_0}}$ while $GL(n,\Bbb C)$ has cardinality $\beth_1=2^{\aleph_0}$, so there aren't even any injective functions from $G$ to $GL(n,\Bbb C)$.