What is a twisted symmetric group?

I am trying to calculate the total number of subgroups for each subgroup in $S_5$. I am working on subgroups of order $6$.

From this website, I have stumbled upon a subgroup called "twisted $S_3$". I understand its generating set of a representative subgroup in the context of $S_5$: it is a $3$-cycle in $S_5$, and a double transposition constructed through selecting two elements in the $3$-cycle and two elements not in the $3$-cycle (hence why there are $\frac{{5 \choose 2}2!}{2}\frac{{3 \choose 2}1!}{3}\frac{{2 \choose 2}1!}{1}= 10$ elements).

What is the twisted symmetric group in general, though?

Solution 1:

I believe "twisted" is being used in a loose sense here. Take a mathematical object, split it into pieces, augment each piece (in a way that is evocative of "twisting") in a way that varies meaningfully with the choice of piece (e.g. smoothly or homomorphically), then put the pieces back together: now you have a twisted version of your original mathematical object. I think this applies to Fourier analysis and tensor products of vector bundles, for example. You could say the Mobius band is a twisted version of a band by this definition, for another example.

In this context, if $H\le G$ is a subgroup and $\phi:H\to C_G(H)$ a homomorphism, then

$$ \{h\phi(h)\mid h\in H\} $$

is an isomorphic copy of $H$ that has been "twisted" by $\phi$ within $G$.

For another example with finite groups, I explained here how the binary octahedral group $2O$ is isomorphic to a "fake $\mathrm{GL}_2\mathbb{F}_3$" (read: twisted version within $\mathrm{GL}_2\mathbb{F}_{3^2}$). You have to interpret "twisting" slightly looser to understand that example, since $\phi$ is really only a homomorphism to $\mathrm{PGL}_2\mathbb{F}_{3^2}$ there, which is why the result is not actually isomorphic to $\mathrm{GL}_2\mathbb{F}_3$.