Decomposition of a cycle as a product of transpositions

Can someone please explain the rules pertaining to different ways to write a cycle decomposition as products of 2-cycles, an example from textbook: I understand this $$(12345) = (54)(53)(52)(51) ,$$ but it also can be written $$(12345) = (54)(52)(21)(25)(23)(13)$$ which I don't understand how it can be.

• Have you tried writing out the right hand sides and multiplying them? Oct 13 '13 at 13:36
• Now i understand. Thanks for the comment Oct 13 '13 at 13:48

Usually by a "cycle decomposition" one means writing a permutation as a product of disjoint cycles. So neither of your products of transpositions is a cycle decomposition.

More formally, a "product of disjoint cycles" is a product of the form $\sigma_1\sigma_2\cdots\sigma_n$ for $n\ge 0$, where every $\sigma_i$ is a cycle and $\sigma_i$ is disjoint from $\sigma_j$ unless $i=j$.

The case $n=0$ is the empty product which by definition is the identity element, so $e$.

In the case $n=1$ you have only one cycle, which is the product if itself and nothing else! This product automatically one of "disjoint cycles" in the vacuous sense that $i=j$ whenever $1\le i\le 1$ and $1\le j\le 1$.

You should have a theorem somewhere describing the uniqueness of cycle decomposition, so $(1\,2\,3\,4\,5)$ is itself its own cycle decomposition, and is its only cycle decomposition, except for other ways to write down the same cycle, such as $(3\,4\,5\,1\,2)$.

$(13):(12345)->(32145)$

$(23)(13):(12345)->(23145)$

$(25)(23)(13):(12345)->(53142)$

$(21)(25)(23)(13):(12345)->(53241)$

$(52)(21)(25)(23)(13):(12345)->(23541)$

$(54)(52)(21)(25)(23)(13):(12345)->(23451)$

Since $(12345)$ is identical with $(23451)$,

$(54)(52)(21)(25)(23)(13)=(23451)$

Notice that $\left(54\right)\left(53\right)\left(52\right)\left(51\right)$ is actually a composite of permutations. Permution $\left(153\right)$ for instance sends $1$ to $5$, sends $5$ to $3$ and sends $3$ to $1$. Here $2$ and $4$ (and eventually others) are untouched, or equivalently are sent to themselves. Questioning where $1$ will be sent to by $\left(54\right)\left(53\right)\left(52\right)\left(51\right)$ we find $\left(54\right)\left(53\right)\left(52\right)\left(51\right)\left(1\right)=\left(54\right)\left(53\right)\left(52\right)(5)=\left(54\right)\left(53\right)\left(2\right)=\left(54\right)\left(2\right)=2$. Likewise we find that it sends $2$ to $3$, $3$ to $4$, $4$ to $5$ and $5$ to $1$. So it can be recognized as permutation $\left(12345\right)$. Composite $\left(54\right)\left(52\right)\left(21\right)\left(25\right)\left(23\right)\left(13\right)$ does exactly the same so can also be recognized as that permutation.