I have a question on grap theory as follows
$G=(V,E)$ is a connected graph. Prove that G is Eulerian if and only if there is a partition $E_j$, $j=1,...,m$ of the set of edges such that every $E_j$ is the set of edges of a cycle of $G$.
I would like you to give ideas for solutions. Thank you.
You should note that if every vertex had degree at least $2$, then there exists a cycle.
There is another fact you should be using- if every vertex in a graph has an even degree, then the graph has a decomposition into edge-disjoint cycles. The proof sketch for this is as follows. Take a maximial list of edge-disjoint cycles within the graph $G$. Suppose every vertex in $G$ has an even degree. Now remove all the edges of these cycles from $G$, and suppose $G$ still contains an edge. This implies at least one vertex in $G$ has odd degree, a contradiction.
So if $G$ is Eulerian, every vertex has even degree, and so the forward side follows trivially.
Now for the converse, suppose $G$ has a decomposition into edge disjoint cycles but not every vertex is even. Then you have a contradiction based on the second fact I stated above.