[Math] what is the order of the subgroup of $S_3$

Question

True or false? $S_{3}$ has a subgroup of order $5$.

$S_3$ is the group of all permutations of the numbers $1,2,$ and $3$. An example of a permutation is, for instance, $(32)$, which means switching $3$ and $2$ (so the string $123$ becomes $132$). In the same way $(123)$ transforms the string $123$ to $231$ ($1\to 2\to 3\to 1$).

Now the group $S_3$ contains the elements:
$\{ e, (12),(13),(23),(123),(132)\}$, which has order $6$ as it should ($3!=6$, the order of $S_n$ is $n!$).

Now of course $\{e\}$ and $S_3$ itself are subgroups (they always are: for a group $G$, $\{e\}\le G$ and $G\le G$).
Now another subgroup is $\{e,(123),(132)\}$, for $(132)$ is the inverse of $(123)$, $e$ is in there and the product of any two elements is again in the subgroup.

Now note that the elements $(12),(13)$ and $(23)$ have order $2$–i.e. $(12)(12) = e$ for if we switch $1$ and $2$ and then switch them again we've changed nothing–so they all generate a subgroup of order $2$:

$$\begin{cases}\{e,(12)\}\\
\{e,(13)\}\\
\{e,(23)\}\end{cases}$$

********These are all the subgroups of $S_3$ as far as I know******
*************thank you ******* so the answer is false ***am i right*****

Answer 1

I assume you haven't seen Lagrange's theorem, as that makes this question much easier. As a result, we can hash through this the old-fashioned way.

Since subgroups are closed under inverses, and the $3$-cycles, $(123),(321)$ are inverses of one another, any subgroup of order $5$ must have both of them and the identity (because otherwise it is missing more than one element, and cannot have size $5$), which means if the subgroup exists, then it is just equal to $S_3\setminus\{x\}$ for some $x\in\{(12),(23),(13)\}$. Let us call this candidate set "$H$".

First note that $(123)=(12)(23)$

Case 1: $(12\not\in H$. Then as $(123),(23)\in H$ and $(123)(23)=(12)$ we have a contradiction.

Case 2: $(23)\not\in H$. Then $(12),(123)\in H$ and $(12)(123)=(23)$, a contradiction.

Case 3: $(13)\not\in H$. Then note $(12),(23)\in H$ and $(12)(23)(12)=(13)$ a contradiction, hence there is no way to have a subset of order $5$ which is also a group.

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In the case you have seen Lagrange's theorem and just cannot see how to apply it, we note that subgroups have the property that if $H\le G$ then the order of every element divides the order of the group. But then we know that the identity, $e$ is the only element of order $1$, and we can see directly from examining each element that the orders of all elements of $G$ are one of $1,2$ or $3$. As a result, if we denote by $|x|$ the order of an element $x$, then when $|x|\ne 1$ we must have $|x|\in\{2,3\}$, and neither of these divide $5$, hence there can be no subgroup of that size since non-identity elements of such a subgroup would have to have order dividing $5$, and neither of $2,3$ divide $5$.