What is the Group of symmetries of Texas Holdem Poker hands? In Texas Holdem poker, each player is dealt two cards from a standard deck of 52 playing cards. (2,3,4...Queen,King,Ace). What is the group of all the symmetries of all (two-card) poker hands? (Given suits are equivalent, see below)
I like to envisage a poker hand as an abstract mathematical object, and I'd quite like to visualise the group of symmetries. Obviously we have the rank of the card, effectively 1 to 13, but also we have the four suits. In poker, there is no heirarchy to the suits, a spade is no more, or less, important than a club. So Ace-hearts-King-hearts is the same, in effect, as the Ace-diamonds-King-diamonds. Likewise, Ace-spades-Ace-hearts is, in effect, the same hand as Ace-Spaces-Ace-Diamonds. And so on. [From Wikipedia - "There are 169 non-equivalent starting hands in hold 'em, which is the sum total of : 13 pocket pairs, 13 × 12 / 2 = 78 suited hands and 78 unsuited hands (13 + 78 + 78 = 169)."]
I supose that the group of symmetries of suited hands are a sub-group of the group of symmetries of all hands. Also, the group of symmetries of paired hands are another sub-group of the group of all hands.
I'm an amateur with knowlege of the basics, so, if fancy symbols, or elaborate jargon, are used, please explain them to the physics-degree-educated layman who has an understanding of the non-technical basics of group theory. I know the axioms, for example.
Also, what are the representations (matrix?) of the hands, paired hands, and suited hands?
 A: If you're thinking about the symmetries with respect to suites, you have $S_4$, the symmetric group on $4$ letters. Let $\{S,D,C,H\}$ be the suites. Elements of this group look like $\sigma = ($C,D$)$ which permutes clubs and diamonds in a given hand, and leaves the other suites untouched. It is helpful to think of this really in terms of group actions, with the set of suite permutation $S_4$ acting on the set of hands.
We can use information about $S_4$ to re-derive the 169 number you mentioned in an interesting way. We define an $n$-cycle as a single cycle of length $n$. Any element of $S_4$ has what is known as a cycle structure (which you can read more about here: https://en.wikipedia.org/wiki/Cyclic_permutation). For example $(S,D,C,H)$ has cycle structure $4$ since it is composed of just one $4$-cycle, and $(C,S)(D,H)$ has cycle structure $2-2$ because it is composed of two $2$ cycles.
We can easily check that there are only $5$ possible cycle structures in $S_4$:
$0$ cycle: This is just the identity permutation, leaves all suites unchanged usually denoted ()
$2$ cycle: This just swaps any two suites, e.g $(C,D)$ or $(S,D)$
$3$ cycle: Permutes any three of the four suites, e.g. $(S,H,D)$ or $(S,D,H)$
$4$ cycle: Permutes all of the suites, e.g. $(C,S,D,H)$, or $(C,H,S,D)$
$2-2$ cylce: Permutes two pairs of suites, e.g. $(D,S)(C,H)$ or $(S,H)(D,C)$
We can obtain the number of equivalent hands by counting the number of hands which are fixed by each of these cycles types and dividing by the number of elements in the group, $24$. It's easy to see that the identity permutation fixes every hand giving us $1326$.
Any $2$ cycle is going to fix a hand if either the hand contains neither of the permuted suites of which there are $325$, or if both cards are the same rank and one of each of the permuted suites, which yields an additional $13$, of a total of $338$. But there are also $6$ possible $2-2$ cycles, yielding a total of $2028$ fixed hands.
I won't belabor the computation any further as you might find it interesting, but it should come out to $624$ fixed hands under the $3$ cycles, $78$ under the $2-2$ cycles, and $0$ under the $4$ cycles. This sums to $1326+2028+624+78 = 4056$ and dividing by the order of the group gives us $4056/24 = 169$. This is also nice because it is fairly easy to see how you would apply this to other games.
Addendum:
We can think about the general case here for a moment, and while you can certainly apply these ideas here, I don't think they are insightful. Instead of the $S_4$, we can think about the action of $S_4\times S_{13}$ on the set of all hands, where we define the permutation in the first coordinate to be the permutation of the suite and the second coordinate to be the permutation of the rank. One can easily verify that this is indeed a group action.
We can now realize the suite invariance property we explored above as a subgroup $S_4\times S_{13}$, namely the subrgoup composed of elements of the form $(\sigma,())$ where $()$ denotes the identity permutation as above. My understanding is that rank pretty much always matters, so I'm not sure there is much you can do with the second coordinate. I would love for someone to prove me wrong, but I don't think you're going to get a lot more out of this structure. The problem is that while we can use group actions to study the set of hands, there doesn't seem to be a way to actually form an algebraic structure from the set itself. There is no binary operation with which to compose hands into new ones which is an apriori requirement.
