2
  • Solution is apparently 6.

Current attempt, although, I am not sure whether this the right approach:

Let $f(x):=x$. We require $f(63)=63x\equiv 0 \mod 147$; that is, $63x=147k\iff 7\mid x$. So we have 21 distinct homomorphisms.

If we consider a few, say,

$f_7(a)=7a\in \mathbb Z_{147}\implies |img(f_7)|=\lfloor \frac{147}{7}\rfloor =21$,

$f_{14}(a)=14a\in \mathbb Z_{147}\implies |img(f_{14})|=\lfloor \frac{147}{14}\rfloor =10$,

$\vdots$

$f_{7t}(a)=7ta \implies |img(f_{7t})|=\lfloor \frac{21}{t}\rfloor$ for $t\in\{0,...,20\}$(verify inductively).

Thus, setting $\lfloor \frac{21}{t}\rfloor=7\implies t=3$ as the unique $t$ within our specified range. What other 5 homomorphisms with images of size 7 am I missing? Ideally, I would like a general method for finding any image size

Anne Bauval
  • 34,650
uouowo
  • 21
  • 2
  • 1
    There is only one subgroup of order $7$ in $\Bbb Z_{147}$ and this subgroup is generated by any of its 6 nonzero elements. – Anne Bauval Mar 02 '24 at 06:13
  • How did you get that there is only one subgroup of order 7 in \mathbb Z_{147}? – uouowo Mar 02 '24 at 06:45
  • The kernel has order $9$; the elements of each of the six nontrivial cosets of the kernel are all sent to one same of the six generators of the image. Once you have done it with one coset, the images of the others are determined. You can do this routine in six different ways. – citadel Mar 02 '24 at 06:51
  • @AnneBauval also, why can't I conclude $7\mid x$? $63x=147k \iff 3x=7k$, so $7\mid 3x\implies$ Euclid's lemma $\implies 7\mid x$? – uouowo Mar 02 '24 at 06:56

2 Answers2

2

Short solution: there is only one subgroup of order $7$ in $\Bbb Z_{147}$ and since $7$ is prime, this subgroup is generated by any of its $6$ nonzero elements, so there are exactly $6$ possible choices for your $x:=f(1)$.

Solution following (and correcting) your approach: a homomorphism $f:\mathbb Z_{63}\to\mathbb Z_{147}$ is determined by the choice of $x:=f(1)\in\mathbb Z_{147}$ such that $7\mid x$. Let us denote it (like you did) by $f_{7t}$, where $x=7t$ and $t\in\{0,1,\dots,20\}$. Then, $|img(f_{7t})|$ is not equal to your $\lfloor\frac{21}{t}\rfloor$ (for instance $|img(f_{14})|$ cannot be $10$ since it must divide $147$) but to $\frac{21}{\gcd(21,t)}$, so that it equals $7$ iff $\gcd(21,t)=3$. This leaves us with $6$ solutions: $t\in\{3,6,9,12,15,18\}$.

More generally, by the same reasoning, for any of the four divisors $d=1,3,7,21$ of $21$, the number of homomorphisms $f:\mathbb Z_{63}\to\mathbb Z_{147}$ such that $|img(f)|=d$ is the number of integers $t\in\{0,1,\dots,20\}$ such that $\gcd(21,t)=\frac{21}d$, i.e. respectively $1,2,6,12$ (you may check that $1+2+6+12=21$).

Anne Bauval
  • 34,650
1

There are a number of homomorphisms from $\Bbb Z_{63}$ to $\Bbb Z_{147}. $

As usual since $\Bbb Z_{63}$ is cyclic, each is determined by $h (1).$

The only requirement is that $\mid h (1)\mid\mid63.$

A cyclic group has one cyclic subgroup of every order dividing its order. There's $\varphi (7)=6$ generators for the subgroup of order $7$ in $\Bbb Z_{147}.$

Defining $h(1)$ to be any of these six gives different homomorphisms with image order $7.$

For an image size $m$, there's $\varphi (m)$ generators to choose from, where $\varphi $ is Euler's phi function.

Any of the mutual divisors of $147$ and $63$ is a possibility. So, $m=1,3,7,$ or $21.$

The number of homomorphisms in each case is $1,2,6$ or $12.$

calc ll
  • 8,427