Prove that a quotient group has an element of order 2. 
Let $H$, $K$ be subgroups of $G$, with $H$ being a normal subgroup of $G$. Suppose the cardinality of $H\cap K$ is odd and $K$ has an element $k$ of order $2$. How would you prove that the quotient group $G/H$ has an element of order $2$? 

I really don't understand this material well, would it be possible to be explained at an elementary level?
 A: Step 1. Show that $H\cap K$ is a subgroup
Step 2: Show that $k$ cannot be in $H\cap K$
Step 3: Consider $k H \in G/H$
A: If $k\in K$ has order two, this means that $k^2 = e$. $k$ cannot belong to $H$, because $H\cap K$ is a subgroup of $G$, and the order of any element in a group divides the order of the group (Lagrange's theorem). If $H\cap K$ has an odd number of elements, then there can be no element of even order in $H\cap K$. Elements of the quotient group are of the form $gH$ for $g\in G$, so consider $kH$. Because $k\not\in H$, $kH\neq eH$, so $kH$ is a nonidentity element of $G/H$. But $(kH)^2 = k^2 H = eH$, so $kH$ has order two, which is what was to be shown.
A: Here's a rare case where we can use the full power of the second isomorphism theorem. 
We prove that $HK/H$ has an element of order $2$. By the second isomorphism theorem, this is isomorphic to $K/(H\cap K)$. Since $K$ has an element of order $2$, $|K|$ is even by Lagrange's theorem. By assumption $|H\cap K|$ is odd, so $|K/(H\cap K)|=|K|/|H\cap K|$ is even. By Cauchy's theorem there is an element of order $2$, hence there is an element of order $2$ in $HK/H\subseteq G/H$.
