Which of the following sets form a group under mutliplication modulo $14$? Which of the following sets form a group under mutliplication modulo 14?


*

*$\{1,3,5\}$

*$\{1,3,5,7\}$

*$\{1,7,13\}$

*$\{1,9,11,13\}$


I figured that only $\{1,3,5\}$ forms a group. But my answers say that that one is also wrong because 1) $3x3=9$ so $9$ should be in there as well and 2) $5\times5=11 \mod 14$ and $5$ is in there but $11$ is not. Could someone explain this reasoning to me please? 
 A: The invertible residue classes with respect to multiplication modulo $14$ are $1,3,5,9,11,13$, according with the fact that
$$
\varphi(14)=\varphi(2)\varphi(7)=6
$$
This immediately excludes 2 and 4, because subsets of a finite group that are closed under multiplication are subgroups.
For 1, consider $3^2=9$.
For 3, note that $7$ is not invertible.

You can also consider the fact that proper subgroups can only have order $1$, $2$ or $3$. Since
$$
3^2=9,\quad 3^3\equiv 13\pmod{14}
$$
you know that $3$ has order $6$, so it's a generator of the whole group. Therefore the only proper subgroups are
$$
\{1\}, \{1,13\}, \{1,9,11\}
$$
A: For one of the sets to be a group, you need to check if any multiplication of set elements (modulo 14) is again in the set and every element is invertible. More formally:
$S$ is a group if and only if the following hold:


*

*for all $a,b \in S$ it is true that $a \cdot b \mod 14 \in S$

*for all $a \in S$ there exists a $b \in S$ such that $ab = 1$.


The set $\lbrace 1, 7, 13 \rbrace$ is closed under multiplication, but $7$ has no inverse element: $7 * 7 \equiv 7 \mod 14$ and $7 * 13 \equiv 7 \mod 14$.
For the first two sets you already have the answer.
For the last set you get $9 * 13 \mod 14 \equiv 5$ which is not in $\lbrace 1, 9, 11, 13\rbrace$.
A: $3\cdot 3=9\equiv_{14}9\notin\{1,3,5\}$;
$3\cdot 3=9\equiv_{14}9\notin\{1,3,5,7\}$;
$9\cdot 13=117\equiv_{14}5\notin\{1,9,11,13\}$.
We showed that the sets labelled as 1, 2 and 4 are not closed under multiplication mod 14, this is they are not groupoids, hence they cannot be groups.
Instead, $a\cdot b\mod 14\in\{1,7,13\}$, for all $a,b\in\{1,7,13\}$, this is set number 3 is closed under the operation. Inverse elements here:
$1\cdot 1=1\equiv_{14}1$ so $1$ is the inverse of itself;
$13\cdot 13=169\equiv_{14}1$ so $13$ is the inverse of itself.
But
$${7\cdot 1\not\equiv_{14}1,}\qquad{7\cdot 7\not\equiv_{14}1,}\qquad{7\cdot 13\not\equiv_{14}1,}$$
indeed $7$ has no inverse $\mod 14$ and $\{1,7,13\}$ is neither a group.
