Intereting Posts

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For every $g\in G$ there exists an $h\in G$ such that $g = h^3$
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Abelian $2$-groups

So I’m given the following definition:

$h(g)p(z)=p(g^{-1}z)$ where g is an element of $SL(3,\mathbb{C})$, $p$ is in the vector space of homogenous complex polynomials of $3$ variables and $z$ is in $\mathbb{C}^3$.

What I’m having trouble showing is that mapping $g$ to $h(g)$ is a group homomorphism. Namely, I know that $h(ab)p(z)=p(b^{-1}a^{-1}z)$, but I can’t seem to make sense that $h(a)(h(b)(p(z))$ is not $p(a^{-1}b^{-1}z)$.

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- Homomorphsim from a finite group into a divisible abelian group.

This is probably trivial, so condescending replies are welcome!

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You should really write $h(g) p(z)$ as $(h(g) p)(z)$. In other words, $p$ is a polynomial function on $\mathbb{C}^3$ and so is $h(g) p$, with its value at $z$ being given by your formula, i.e. $(h(g) p)(z) = p(g^{-1} z)$.

Now recompute $(h(a) h(b) p)(z)$, giving $(h(a) h(b) p)(z) = (h(b) p)(a^{-1} z) = p(b^{-1} a^{-1} z)$. For what it is worth, this may well be classified as “trivial”, but I found keeping straight what acts on what and how was a significant stumbling block for me when I first studied representation theory and invariant theory. It’s easy to get actions wrong and have calculations fall apart due to these “trivial” issues, so they are most definitely worth straightening out.

You can see also this thread for further explanation in a slightly more abstract setting.

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