Usbrth

This is an extra problem that my Algebra teacher suggested working through for practice.

Let $G$ be a group and let $A subseteq G$. Assume that $K = aAmid a in G $ form a partition of G. Show that there exists $a_0in G$ such that H = $a_0$A is a subgroup of G and $aAmid a in G = bHmid b in G$.

I know that if $K$ is a partition of $G$, then either $aA = bA$ or $aA cap bA = emptyset $ for all $a,b in G$ and that there must exist $a_0in A$ such that for $x in G$, $a_0a = x$ which is equivalent to $a_0 = xa^-1$, since $G$ is a group. This format is similar to $H = a_0A$. I'm struggling with forming this subgroup $H$ and showing it's a subgroup. As for the second part, I think I could just show that each set is a subset of the other.

I am new to Abstract Algebra and struggle with the concepts. I am looking for very straightforward, simple suggestions. Thank you!

Hint: Any partition $P$ of $G$ defines an equivalence relation $sim_P$ (and vice versa). Consider $$[e]_sim_K=gin Gmid gsim_K e,$$ where $e$ is the identity of $G$.

Note: $e$ is not necessarily in $A$.

Here’s the trick we want to use:

Let $gin G$ be arbitrary. Then $gin H$ if and only if $gH = H$ if and only if $gH cap H ne emptyset$.

You should prove this! Use the fact that the translates of $H$ form a partition of $G$.

Now, the first step in proving $H$ is a subgroup is showing that for any $hin H$, its inverse $h^-1$ is also in $H$. Explain why $ein h^-1H$, and why this proves that $h^-1in H$.

The second step in proving $H$ is a subgroup is showing that for any $g,hin H$, their product $gh$ is also in $H$. Using the step above, prove that $gin (gh)H$ and explain why this means that $ghin H$.