# Subgroups of a permutation group

The permutation group $S_{4}$ is defined as the group of all possible permutations of [1234].

i) Find the number of subgroups of $S_{4}$ that have order 2.

ii) A: { [1234], [2143], [3412], [4321] } and B: { [1234], [1243], [2134], [2143] }. Which of A and B are subgroups of $S_{4}$?

Trying to teach myself a Further Maths module on Groups is proving difficult when none of my teachers know the syllabus, any help would be appreciated! Thanks.

• Note: people have different conventions for writing permutations. – Mark Bennet Mar 6 '15 at 19:34

i) You want all transpositions (ab), and all pairs of disjoint transpositions (ab)(cd).

ii) Assuming that [2143] means, in cycle notation, (12)(34), then A is a subgroup.

Hints:

i) If a subgroup has order two then there are exactly one trivial and one non-trivial element. Furthermore the nontrivial element $a\in G$ must fulfill $a^2=e$. So just check with of the elements of $S_4$ fulfill this requirement.

ii) Just check the subgroup axioms for $A$ and $B$.

• I understand that the subgroups of order 2 must contain the identity element, and a self inverse element, but I am struggling to find the general form for a self inverse element, I have found specific examples but there must be a deductive or analytical method to find them rather than exhaustively trying each one? Thanks for the reply. – talfred Mar 6 '15 at 18:51
• Actually you can see it immediately. For instance if $\sigma=(1234)$ is given, then this can't be an element of order 2, because $\sigma$ maps $1$ to $2$ and $2$ to $3\neq 1$. If we consider $\sigma=(12)$ then this is of course an element of order 2. (Transposition). The element $(12)(34)$ is just 2 transposition, hence of order 2. You can do similar methods for the other elements of $S_4$. Does this help you? – Marm Mar 6 '15 at 18:58
• The textbook I am using has very different notation, [4123] for example means that 1->4, 2->1, 3->2, 4->3. Why does the order of the specific element matter? Sorry if I seem ignorant – talfred Mar 6 '15 at 19:15
• What you call a "self-inverse element" is the same as an element of order $1$ or $2$. If $a = a^{-1}$, then by multiplying both sides by $a$ you get $a^2 = e$. Elements that satisfy the latter are exactly the elements of order $1$ or $2$. – Leppala Mar 10 '15 at 9:51

(i) The order of a permutation is easy to compute given its decomposition as a product of disjoint cycles: it is the l.c.m. of the orders of the cycles, i.e. the l.c.m. of the lengths of the cycles.

Hence a permutation of order $2$ is a product of disjoint transpositions. In $S_4$ it the product of at most $2$ disjpoint transpositions.

We have $\dbinom 42=6$ transpositions and as many products of disjoint transpositions.

(ii) If the notation is that of $4$-cycles, neither $A$ nor $B$ are subgroups.

Indeed, in $A$, let $\gamma=(1234),\enspace \gamma'=(2143)$. Then $\,(4321)=\gamma^{-1}=\gamma^3, \enspace (3412)= \gamma'^{-1}= \gamma'^3$. However $\gamma$ and $\gamma'$ have order $4$. If $A$ were a subgroup, it should contain $\gamma^2=(13)(24)$.

For $B$, let $\gamma$ be as above and $\gamma'=(1243)$. Then $B$ contains $\gamma^{-1}$ and $\gamma'^{-1}$, but doesn't contain $\gamma^2$, nor $\gamma'^2$.