Undergrad Group and Identity: Proving (12)(34)² = (12)(34)

[LCSphysicist]

TL;DR

The question is simple, why does "(12)(34) (elements of A4) squares to the identity"?

I am probably missing a crucial point here, but what does it means that (12)(34) squares to the identity? How do we prove it?
((12)(34))² = (12)(34)(12)(34) = (12)(12)(34)(34) = (12)(34) $\neq I$
Is not this the algorithm?

 

[fresh_42]

$(12)(12)=(1)$ so your conclusion is wrong.

You read this from right to left as:
$2\longrightarrow 1 \longrightarrow 2$ and $1\longrightarrow 2 \longrightarrow 1$ so applying both results in the identity. Permutations are functions: $f_{(12)} : \{1,2\} \longrightarrow \{1,2\}$ with $f_{(12)}(1)=2$ and $f_{(12)}(2)=1$. Multiplication of these functions is applying both functions subsequently.

 

[member 587159]

((12)(34))² = (12)(34)(12)(34) = (12)(12)(34)(34) = (12)(34) $\neq I$

First equality is ok.
Second equality is ok, but it should be justified why $(12)$ and $(34)$ commute.
Third equality is wrong: $(12)(12) = 1$ and $(34)(34) = 1$. The reason for this is that $1 \mapsto 2 \mapsto 1$ and $2 \mapsto 1 \mapsto 2$. It doesn't matter here whether you read from left to right or from right to left.