If $H_a $, $H_b$, $H_c$ are half turns, prove that $ H_aH_bH_c=H_cH_bH_a$ Prove: $H_aH_bH_c=H_cH_bH_a$ where $H_a, H_b$, and $H_c$ are half turns about a, b,and c.
How do I show this? The only thing I know is that the product of two half turns is a translation.
 A: I'm assuming you are talking about the planar Euclidean case. Observe that $H_aH_a = H_a^2 = I$ where $I$ is the identity transformation, because two half-turns equal one full turn, i.e. identity. Simply multiply the relation $H_aH_bH_c=H_cH_bH_a$ with $H_a$ on both sides from he right (because $H_a$ is an invertable transformation) and get the equivalent relation $$H_aH_bH_cH_a=H_cH_bH_aH_a = H_cH_b$$ After that apply $H_b$ on both sides from the right and then do the same with $H_c$. As a result you get the equivalent  identity $H_aH_bH_cH_aH_bH_c = I$. If you prove that $H_aH_bH_cH_aH_bH_c = I$ then by going backwards, you have proven that $H_aH_bH_c=H_cH_bH_a$. 
Now, the product $H_aH_bH_cH_aH_bH_c$ splits into $(H_aH_b) \, (H_cH_a) \, (H_bH_c)$. The composite transformation $H_aH_b$ is a translation with translation vector $2(a-b)$. Similarly, $H_cH_a$ is a translation with translation vector $2(c-a)$. However, two translations commute, meaning the order of their application is irrelevant, which means that $(H_aH_b) \ (H_cH_a) = (H_cH_a) \, (H_aH_b)$. The latter identity applied to the product $(H_aH_b) \, (H_cH_a) \, (H_bH_c)$  leads to the identity
$$(H_aH_b) \, (H_cH_a) \, (H_bH_c) = (H_cH_a) \, (H_aH_b) \, (H_bH_c)$$  Remove the parenthesis, as the product is associative, obtaining the chain of identities 
\begin{align}
H_aH_b H_cH_a H_bH_c & = (H_aH_b) \, (H_cH_a) \, (H_bH_c)\\ 
&= (H_cH_a) \, (H_aH_b) \, (H_bH_c) \\
&= H_cH_a H_aH_b H_bH_c\\
&= H_c\,(H_a H_a)\,(H_b H_b)\,H_c\\
&= H_c\,I\,I\,H_c  = H_cH_c\\
&=I
\end{align}
