While reading Dummit and Foote's Abstract Algebra, he briefly mentions that any transposition $(i \, j)$ can be written as $$(i \, j) = \lambda(1 \, 2)\lambda,$$ where $\lambda$ is the permutation that 'interchanges $1$ and $i$, interchanges $2$ and $j$, and leaves all other numbers fixed (if $i = 1$ or $j = 2$, then $\lambda$ fixes $i$ or $j$ respectively)'. I don't understand what it is that $\lambda$ does, and hence cannot prove the above expression.
So does $\lambda$ take $1$ to $i$ and $i$ to $1$ only if $i$ is not already $1$? What does the condition in the bracket mean? Could someone explain it with an example?
That basically means $\lambda=(1i)(2j)$.
For example, if $i=4$ and $j=5$, then $\lambda=(14)(25)$ and notice $$(45)=(14)(25)(12)(14)(25).$$
But if, for example, $i=1$ and $j=5$, then $\lambda=(25)$ and notice $$(15)=(25)(12)(25).$$
The parenthetical comment simply addresses the fact that we wouldn't write $\lambda=(11)(25)$ in this example.