Let $G$ be a group, $p$ a prime dividing $|G|$ and $X = \{(x_0,..., x_{p−1})) ∈ G_p:∏_i x_i = 1\}.$ Let $E$ be the relation defined on $X$ by $(x_0, ..., x_{p−1})E(y_0,..., y_{p−1})$ if there exists $k ∈ \Bbb Z/p\Bbb Z$ such that $y_i+k = x_i$. (The indexes are considered to live in $\Bbb Z/p\Bbb Z$, so if $i+k⩾p$, what we mean is $y_{i+k−p} = x_i$).
- Show that $|X| = |G|^{p−1}$.
- Show that $E$ is an equivalence relation.
- Show that an equivalence class of $E$ contains a single element if and only if this element is of the form $(x,..., x)$ where $x_p = 1$.
- Show that the equivalence classes of $E$ that are not singletons have cardinal $p$.
- Show that $p$ divides the number of equivalence classes of $E$ that are singletons.
- Show that there exists an element $x ∈ G$ of order $p$.
I'm pretty confused as to where to start. Some things in the problem look familiar. I think $\Bbb Z/p\Bbb Z$ seems important because these are groups with only trivial subgroups. $p$ divides $|G|$ so G has an order $np$ where $n$ is an integer. $X$ is a subset of $G$ in which all of the elements multiplied by each other equals 1. I'm confused about the part where E is defined (is it assuming that there are two subsets of $G$, $X$ and $Y$, that fulfill the product = 1 requirement? And with that assumption, E is only true if there is a $k$ mod $p$ such that for all $y_i$ and its corresponding $x_i$, $y_i+k = x_i$, which means that $Y$ is just $X$ shifted mod $p$?
Any help as to how to prove these questions would be truly appreciated. Thanks!
I believe your misunderstanding the problem.
For start the set $X$ is not subset of $G$, as you say in the question, indeed the elements of $X$ are $n$-tuples of elements of $G$, hence $X \subseteq G^n$.
That said point $1$ is just a basic combinatorial problem: you have to count in how many way you can buil a $n$-tuple $(x_0,x_1,\dots,x_{n-1})$ such that $x_0 x_1 \dots x_{n-1}=1$ in $G$.
Before to going further it could be interesting observing that $E$ identifies two $n$-tuples if and only if they are obtained one from each other by a cyclic permutation of the components in the $n$-tuple. Related to this there is the remark that for every $(x_0,\dots,x_{n-1})$ such that $x_0 \dots x_{n-1}=1$ you have that also $x_{(i+k\ \mod n)}\dots x_{(n-1+k \mod n)}=1$.
Proving that $E$ is an equivalence relation require to prove that $E$ is reflexive, symmetric and transitive.
The other points are pretty straightforward, try to continue from this and in case you find other problems ask to question.