Two element generating sets that produce a group

Question

We have an additive group of integers modulo $n$, take for example $\Bbb{Z}_{154}$, being the set of integers between $0$ and $153$.

How can I find one pair $(x,y)$ of elements that both belong to $\Bbb{Z}_{154}$, which satisfy the condition that both $\langle x\rangle$ and $\langle y\rangle$ are proper subgroups of $\Bbb{Z}_{154}$, and that $\langle x,y\rangle = \Bbb{Z}_{154}$. What would be the process behind this?

Knowing what set will come from a single element generating set is straightforward enough, however, how do we know what set will be produced by a generating set containing two elements such as $\langle x,y\rangle$?

Answer 1

Since $\langle x\rangle,\langle y\rangle$ are proper subgroups of $\mathbb Z_{154}$, $x$ and $y$ must be not coprime to $154$. Since $\langle x,y\rangle$ is the whole group, $\gcd(x,y)$ must be relatively prime to $154$ – it may be $1$, for example.

Now $154=2×7×11$. We can take $x=2,y=7$.

Answer 2

If you view the cyclic group of order $n$ as consisting of the integers $0,1,\ldots,n-1$ with modulo $n$ addition, then the element $a$ generates a subgroup of order $n/\gcd(a,n)$. In particular, $a$ generates the whole group if and only if $\gcd(a,n)=1$. So any elements that are not relatively prime to $n$ generate a proper subgroup.

On the other hand, if $\langle x,y\rangle$ is the whole group, then it must be possible to obtain $1$ as a linear combination of $x$ and $y$ modulo $n$: that is, it must be possible to find integer $a$ and $b$ such that $1\equiv ax+by\pmod{n}$. That means there is an integer $k$ such that $nk=(ax+by)-1$, or equivalently, that you can find integers $a$, $b$, and $c$ such that $$1 = ax+by+cn.$$ Now, any element of the form $ax+by+cn$ is a multiple of $\gcd(x,y,n)$. So we need $\gcd(x,y,n)=1$ for this to work.

In summary, you want integers $x$ and $y$, $0\leq x,y\lt 154$, such that:

1. $\gcd(x,154)\gt 1$;
2. $\gcd(y,154)\gt 1$;
3. $\gcd(x,y,154)=1$.

Now go at it. Note that the last one can be achieved if either $\gcd(x,y)=1$, or if $\gcd(x,154)$ and $\gcd(y,154)$ are relatively prime, even if $x$ and $y$ are not. So there should be lots of choices....

Answer 3

OP's question:

how do we know what set will be produced by a generating set containing two elements such as $\langle x,y\rangle$?

It is known that

A subgroup of a cyclic group is cyclic

So we have cyclic groups and some relations,

$\quad \langle x \rangle \subset \langle x,y\rangle \subset \Bbb{Z}_{154}$,

$\quad \langle y \rangle \subset \langle x,y\rangle \subset \Bbb{Z}_{154}$,

and using group theory definitions we can write

$\quad \langle x,y\rangle = \Bbb{Z}_{154} \; \text{ iff } \; \exists k \in \langle x,y\rangle \text{ such that the order of } k \text{ is equal to } 154$