Determine whether the following permutation is even or odd and write it as a product of transpositions in two different ways.
$(1527)(3567)(273)$
So far, I have the following: $(1527)(3567)(273) = (15)(12)(17)(35)(36)(37)(27)(23) = (57)(51)(56)(53)$
I can understand the part which states that $(1527)(3567)(273) = (15)(12)(17)(35)(36)(37)(27)(23)$, but I can't understand why that might be equal to $(57)(51)(56)(53)$.
Can someone explain why it is so?
It looks like you're using the fact that $(a_1a_2\ldots a_n)=(a_1a_2)(a_1a_3)\ldots (a_1a_n)$. This leads to the first equality immediately by using the above on each component of the permutation.
To get the second, we first multiply each component out to write the permutation as a cycle $(16357)$ which is equal by cyclically permuting the elements to $(57163)$. We may now use the above decomposition rule again to get that it is equal to $(57)(51)(56)(53)$.
As a tip, there is also another way to decompose a chain into transpositions using the fact that $(a_1a_2\ldots a_{n-1}a_{n})=(a_{n}a_{n-1})(a_{n-1}a_{n-2})\ldots (a_3a_2)(a_2a_1)$ which in the above example would give you $(16357)=(75)(53)(36)(61)$.
I can understand the part which states that $$(1527)(3567)(273) = (15)(12)(17)(35)(36)(37)(27)(23)$$ but I can't understand why that might be equal to $(57)(51)(56)(53)$.
To see why they are equal, you might find it helpful to first write your permutation as a product of disjoint cycles.
So, for example, assuming you are composing products of permutation from left to right, then note that $\varphi = (1527)(3567)(273)$ is not a product of disjoint cycles: we see that $7$ appears in all three of the factors, $2$, $3$ and $5$ each appear twice, etc.
So, composing from left to right gives us $$\varphi = (1527)(3567)(273) = (16357)(2) = (16357) = (63571) = (35716) = (57163) \ldots$$ Then writing the last as the product of transpositions gives us $$(57163) = (57)(51)(56)(63)$$
But recall, the decomposition of permutations into the product of transpositions is NOT unique: What is guaranteed is that for a given permutation, it can always and only be decomposed into the product of an even number of transpositions (hence is an even permutation), or an odd number of transpositions (in this case we call it an odd permutation.)
For your posted permutation, note that:
$$(1527)(3567)(273) = \underbrace{(15)(12)(17)(35)(36)(37)(27)(23)}_{\large 8\;\text{ transpositions: even}} = \underbrace{(57)(51)(56)(53)}_{\large 4\;\text{transpositions: even}}$$
Hence, your permutation is an even permutation.
To answer the first question it is not really necessary to write the permutation as a product of $2$-cycles. $4$-cycles are odd and $3$-cycles are even. More generally $2n$-cycles are odd and $2n+1$-cycles are even. You have odd-odd-even here. This results in an even permutation (the composition of two odds is even, the composition of two evens is even, the composition of odd and even is odd).
Since you write $(1527)=(15)(12)(17)$, I assume that you compose permutations “going to the right”. Thus $$ \sigma=(1527)(3567)(273)=(16357) $$ Thus $\sigma=(16)(13)(15)(17)$; but you can also write $\sigma=(63571)$ or $\sigma=(57163)$, can't you?
