Homomorphism $x^i mapsto y^i$ and the order of $x$ and $y$












1












$begingroup$


I am trying to answer the following question:




Let $G$ and $H$ be two cyclic groups generated by $x$ and $y$
respectively. Determine the condition on the orders $m$ and $n$ of $x$ and
$y$ so that the map $f(x^i)= y^i$ is a homomorphism.




How I attempted to solve the question in the following way:
Let $r$ and $s$ be two arbitrary integers. Then
$$f(x^rx^s) = f(x^{r+s})=y^{r+s}=y^ry^s=f(x^r)f(x^s)$$
Since we made no assumptions about the orders of $x$ and $y$, $f$ is an homomorphism from $G$ to $H$ no matter what the orders of $x$ and $y$ are.



I don't feel confident with my proof and I would love for someone to check it. Finally, I am aware there are some theorems about the order elements of a groups but I am supposed to solve the problem without them.






abstract-algebra group-theory finite-groups







share|cite|improve this question











$endgroup$












  • $begingroup$
    In order to show that $f$ is a homomorphism, you need to show that $f(x_1x_2)=f(x_1)f(x_2)$ for all $x_1, x_2in G$, not just for $x_1, x_2$ equal to powers or some particular element of $G$.
    $endgroup$
    – Steve Kass
    Jan 20 at 1:23










  • $begingroup$
    But isn't it true that because $G=langle x rangle$ all the elements in $G$ are powers of a particular element of $G$ (namely $x$), and therefore proving $f(x^rx^s) = f(x^r)f(x^s)$ for arbitrary $r$ and $s$ is equivalent to prove it for that $f(x_1x_2)=f(x_1)f(x_2)$ for all $x_1,x_2 in G$?
    $endgroup$
    – Lazarus
    Jan 20 at 1:31












  • $begingroup$
    It’s not true that $G=langle xrangle$ for every $xin G$. (In my first comment, I misspelled “of” as “or.”)
    $endgroup$
    – Steve Kass
    Jan 20 at 1:36


















1












$begingroup$


I am trying to answer the following question:




Let $G$ and $H$ be two cyclic groups generated by $x$ and $y$
respectively. Determine the condition on the orders $m$ and $n$ of $x$ and
$y$ so that the map $f(x^i)= y^i$ is a homomorphism.




How I attempted to solve the question in the following way:
Let $r$ and $s$ be two arbitrary integers. Then
$$f(x^rx^s) = f(x^{r+s})=y^{r+s}=y^ry^s=f(x^r)f(x^s)$$
Since we made no assumptions about the orders of $x$ and $y$, $f$ is an homomorphism from $G$ to $H$ no matter what the orders of $x$ and $y$ are.



I don't feel confident with my proof and I would love for someone to check it. Finally, I am aware there are some theorems about the order elements of a groups but I am supposed to solve the problem without them.






abstract-algebra group-theory finite-groups







share|cite|improve this question











$endgroup$












  • $begingroup$
    In order to show that $f$ is a homomorphism, you need to show that $f(x_1x_2)=f(x_1)f(x_2)$ for all $x_1, x_2in G$, not just for $x_1, x_2$ equal to powers or some particular element of $G$.
    $endgroup$
    – Steve Kass
    Jan 20 at 1:23










  • $begingroup$
    But isn't it true that because $G=langle x rangle$ all the elements in $G$ are powers of a particular element of $G$ (namely $x$), and therefore proving $f(x^rx^s) = f(x^r)f(x^s)$ for arbitrary $r$ and $s$ is equivalent to prove it for that $f(x_1x_2)=f(x_1)f(x_2)$ for all $x_1,x_2 in G$?
    $endgroup$
    – Lazarus
    Jan 20 at 1:31












  • $begingroup$
    It’s not true that $G=langle xrangle$ for every $xin G$. (In my first comment, I misspelled “of” as “or.”)
    $endgroup$
    – Steve Kass
    Jan 20 at 1:36
















1












1








1


1



$begingroup$


I am trying to answer the following question:




Let $G$ and $H$ be two cyclic groups generated by $x$ and $y$
respectively. Determine the condition on the orders $m$ and $n$ of $x$ and
$y$ so that the map $f(x^i)= y^i$ is a homomorphism.




How I attempted to solve the question in the following way:
Let $r$ and $s$ be two arbitrary integers. Then
$$f(x^rx^s) = f(x^{r+s})=y^{r+s}=y^ry^s=f(x^r)f(x^s)$$
Since we made no assumptions about the orders of $x$ and $y$, $f$ is an homomorphism from $G$ to $H$ no matter what the orders of $x$ and $y$ are.



I don't feel confident with my proof and I would love for someone to check it. Finally, I am aware there are some theorems about the order elements of a groups but I am supposed to solve the problem without them.






abstract-algebra group-theory finite-groups







share|cite|improve this question











$endgroup$




I am trying to answer the following question:




Let $G$ and $H$ be two cyclic groups generated by $x$ and $y$
respectively. Determine the condition on the orders $m$ and $n$ of $x$ and
$y$ so that the map $f(x^i)= y^i$ is a homomorphism.




How I attempted to solve the question in the following way:
Let $r$ and $s$ be two arbitrary integers. Then
$$f(x^rx^s) = f(x^{r+s})=y^{r+s}=y^ry^s=f(x^r)f(x^s)$$
Since we made no assumptions about the orders of $x$ and $y$, $f$ is an homomorphism from $G$ to $H$ no matter what the orders of $x$ and $y$ are.



I don't feel confident with my proof and I would love for someone to check it. Finally, I am aware there are some theorems about the order elements of a groups but I am supposed to solve the problem without them.






abstract-algebra group-theory finite-groups




abstract-algebra group-theory finite-groups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 20 at 1:21







Lazarus

















asked Jan 20 at 1:06









LazarusLazarus

11217




11217












  • $begingroup$
    In order to show that $f$ is a homomorphism, you need to show that $f(x_1x_2)=f(x_1)f(x_2)$ for all $x_1, x_2in G$, not just for $x_1, x_2$ equal to powers or some particular element of $G$.
    $endgroup$
    – Steve Kass
    Jan 20 at 1:23










  • $begingroup$
    But isn't it true that because $G=langle x rangle$ all the elements in $G$ are powers of a particular element of $G$ (namely $x$), and therefore proving $f(x^rx^s) = f(x^r)f(x^s)$ for arbitrary $r$ and $s$ is equivalent to prove it for that $f(x_1x_2)=f(x_1)f(x_2)$ for all $x_1,x_2 in G$?
    $endgroup$
    – Lazarus
    Jan 20 at 1:31












  • $begingroup$
    It’s not true that $G=langle xrangle$ for every $xin G$. (In my first comment, I misspelled “of” as “or.”)
    $endgroup$
    – Steve Kass
    Jan 20 at 1:36




















  • $begingroup$
    In order to show that $f$ is a homomorphism, you need to show that $f(x_1x_2)=f(x_1)f(x_2)$ for all $x_1, x_2in G$, not just for $x_1, x_2$ equal to powers or some particular element of $G$.
    $endgroup$
    – Steve Kass
    Jan 20 at 1:23










  • $begingroup$
    But isn't it true that because $G=langle x rangle$ all the elements in $G$ are powers of a particular element of $G$ (namely $x$), and therefore proving $f(x^rx^s) = f(x^r)f(x^s)$ for arbitrary $r$ and $s$ is equivalent to prove it for that $f(x_1x_2)=f(x_1)f(x_2)$ for all $x_1,x_2 in G$?
    $endgroup$
    – Lazarus
    Jan 20 at 1:31












  • $begingroup$
    It’s not true that $G=langle xrangle$ for every $xin G$. (In my first comment, I misspelled “of” as “or.”)
    $endgroup$
    – Steve Kass
    Jan 20 at 1:36


















$begingroup$
In order to show that $f$ is a homomorphism, you need to show that $f(x_1x_2)=f(x_1)f(x_2)$ for all $x_1, x_2in G$, not just for $x_1, x_2$ equal to powers or some particular element of $G$.
$endgroup$
– Steve Kass
Jan 20 at 1:23




$begingroup$
In order to show that $f$ is a homomorphism, you need to show that $f(x_1x_2)=f(x_1)f(x_2)$ for all $x_1, x_2in G$, not just for $x_1, x_2$ equal to powers or some particular element of $G$.
$endgroup$
– Steve Kass
Jan 20 at 1:23












$begingroup$
But isn't it true that because $G=langle x rangle$ all the elements in $G$ are powers of a particular element of $G$ (namely $x$), and therefore proving $f(x^rx^s) = f(x^r)f(x^s)$ for arbitrary $r$ and $s$ is equivalent to prove it for that $f(x_1x_2)=f(x_1)f(x_2)$ for all $x_1,x_2 in G$?
$endgroup$
– Lazarus
Jan 20 at 1:31






$begingroup$
But isn't it true that because $G=langle x rangle$ all the elements in $G$ are powers of a particular element of $G$ (namely $x$), and therefore proving $f(x^rx^s) = f(x^r)f(x^s)$ for arbitrary $r$ and $s$ is equivalent to prove it for that $f(x_1x_2)=f(x_1)f(x_2)$ for all $x_1,x_2 in G$?
$endgroup$
– Lazarus
Jan 20 at 1:31














$begingroup$
It’s not true that $G=langle xrangle$ for every $xin G$. (In my first comment, I misspelled “of” as “or.”)
$endgroup$
– Steve Kass
Jan 20 at 1:36






$begingroup$
It’s not true that $G=langle xrangle$ for every $xin G$. (In my first comment, I misspelled “of” as “or.”)
$endgroup$
– Steve Kass
Jan 20 at 1:36












1 Answer
1






active

oldest

votes


















1












$begingroup$

Let $g_1, g_2 in G$. We then have that $g_1=x^{k_1}$ and $g_2=x^{k_2}$ for $k_1,k_2 in mathbb{Z}$. Let $f$ be a homomorphism. Then
$$f(g_1g_2)=f(x^{k_1}x^{k_2})=f(x^{k_1+k_2})=f(g_1)f(g_2)=f(x^{k_1})(x^{k_2})=y^{k_1}y^{k_2}=y^{k_1+k_2}.$$



Further, recall that for any group homomorphism $varphi: G to H$ that for $g in G$ we have that $|varphi(g)|$ divides $|g|$.



Thus from above we must have that $|y^{k_1+k_2}|$ divides $|x^{k_1+k_2}|$. And since $G=langle x rangle$ and $H=langle y rangle$ with $|x|=m$ and $|y|=n$, it clearly follows that $G cong mathbb{Z}/mmathbb{Z}$ and $H cong mathbb{Z}/nmathbb{Z}$. Therefore
$$|y^{k_1+k_2}|=frac{n}{gcd(k_1+k_2,n)},$$
$$|x^{k_1+k_2}|=frac{m}{gcd(k_1+k_2,m)}.$$
Using that the first expression divides the second, we have for $r in mathbb{Z}$
$$|x^{k_1+k_2}|=frac{m}{gcd(k_1+k_2,m)}=|y^{k_1+k_2}|cdot r = frac{nr}{gcd(k_1+k_2,n)}$$
Providing us with
$$m=frac{gcd(k_1+k_2,m)}{gcd(k_1+k_2,n)} cdot nr$$






share|cite|improve this answer









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    1 Answer
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    1 Answer
    1






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    active

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    active

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    1












    $begingroup$

    Let $g_1, g_2 in G$. We then have that $g_1=x^{k_1}$ and $g_2=x^{k_2}$ for $k_1,k_2 in mathbb{Z}$. Let $f$ be a homomorphism. Then
    $$f(g_1g_2)=f(x^{k_1}x^{k_2})=f(x^{k_1+k_2})=f(g_1)f(g_2)=f(x^{k_1})(x^{k_2})=y^{k_1}y^{k_2}=y^{k_1+k_2}.$$



    Further, recall that for any group homomorphism $varphi: G to H$ that for $g in G$ we have that $|varphi(g)|$ divides $|g|$.



    Thus from above we must have that $|y^{k_1+k_2}|$ divides $|x^{k_1+k_2}|$. And since $G=langle x rangle$ and $H=langle y rangle$ with $|x|=m$ and $|y|=n$, it clearly follows that $G cong mathbb{Z}/mmathbb{Z}$ and $H cong mathbb{Z}/nmathbb{Z}$. Therefore
    $$|y^{k_1+k_2}|=frac{n}{gcd(k_1+k_2,n)},$$
    $$|x^{k_1+k_2}|=frac{m}{gcd(k_1+k_2,m)}.$$
    Using that the first expression divides the second, we have for $r in mathbb{Z}$
    $$|x^{k_1+k_2}|=frac{m}{gcd(k_1+k_2,m)}=|y^{k_1+k_2}|cdot r = frac{nr}{gcd(k_1+k_2,n)}$$
    Providing us with
    $$m=frac{gcd(k_1+k_2,m)}{gcd(k_1+k_2,n)} cdot nr$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Let $g_1, g_2 in G$. We then have that $g_1=x^{k_1}$ and $g_2=x^{k_2}$ for $k_1,k_2 in mathbb{Z}$. Let $f$ be a homomorphism. Then
      $$f(g_1g_2)=f(x^{k_1}x^{k_2})=f(x^{k_1+k_2})=f(g_1)f(g_2)=f(x^{k_1})(x^{k_2})=y^{k_1}y^{k_2}=y^{k_1+k_2}.$$



      Further, recall that for any group homomorphism $varphi: G to H$ that for $g in G$ we have that $|varphi(g)|$ divides $|g|$.



      Thus from above we must have that $|y^{k_1+k_2}|$ divides $|x^{k_1+k_2}|$. And since $G=langle x rangle$ and $H=langle y rangle$ with $|x|=m$ and $|y|=n$, it clearly follows that $G cong mathbb{Z}/mmathbb{Z}$ and $H cong mathbb{Z}/nmathbb{Z}$. Therefore
      $$|y^{k_1+k_2}|=frac{n}{gcd(k_1+k_2,n)},$$
      $$|x^{k_1+k_2}|=frac{m}{gcd(k_1+k_2,m)}.$$
      Using that the first expression divides the second, we have for $r in mathbb{Z}$
      $$|x^{k_1+k_2}|=frac{m}{gcd(k_1+k_2,m)}=|y^{k_1+k_2}|cdot r = frac{nr}{gcd(k_1+k_2,n)}$$
      Providing us with
      $$m=frac{gcd(k_1+k_2,m)}{gcd(k_1+k_2,n)} cdot nr$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Let $g_1, g_2 in G$. We then have that $g_1=x^{k_1}$ and $g_2=x^{k_2}$ for $k_1,k_2 in mathbb{Z}$. Let $f$ be a homomorphism. Then
        $$f(g_1g_2)=f(x^{k_1}x^{k_2})=f(x^{k_1+k_2})=f(g_1)f(g_2)=f(x^{k_1})(x^{k_2})=y^{k_1}y^{k_2}=y^{k_1+k_2}.$$



        Further, recall that for any group homomorphism $varphi: G to H$ that for $g in G$ we have that $|varphi(g)|$ divides $|g|$.



        Thus from above we must have that $|y^{k_1+k_2}|$ divides $|x^{k_1+k_2}|$. And since $G=langle x rangle$ and $H=langle y rangle$ with $|x|=m$ and $|y|=n$, it clearly follows that $G cong mathbb{Z}/mmathbb{Z}$ and $H cong mathbb{Z}/nmathbb{Z}$. Therefore
        $$|y^{k_1+k_2}|=frac{n}{gcd(k_1+k_2,n)},$$
        $$|x^{k_1+k_2}|=frac{m}{gcd(k_1+k_2,m)}.$$
        Using that the first expression divides the second, we have for $r in mathbb{Z}$
        $$|x^{k_1+k_2}|=frac{m}{gcd(k_1+k_2,m)}=|y^{k_1+k_2}|cdot r = frac{nr}{gcd(k_1+k_2,n)}$$
        Providing us with
        $$m=frac{gcd(k_1+k_2,m)}{gcd(k_1+k_2,n)} cdot nr$$






        share|cite|improve this answer









        $endgroup$



        Let $g_1, g_2 in G$. We then have that $g_1=x^{k_1}$ and $g_2=x^{k_2}$ for $k_1,k_2 in mathbb{Z}$. Let $f$ be a homomorphism. Then
        $$f(g_1g_2)=f(x^{k_1}x^{k_2})=f(x^{k_1+k_2})=f(g_1)f(g_2)=f(x^{k_1})(x^{k_2})=y^{k_1}y^{k_2}=y^{k_1+k_2}.$$



        Further, recall that for any group homomorphism $varphi: G to H$ that for $g in G$ we have that $|varphi(g)|$ divides $|g|$.



        Thus from above we must have that $|y^{k_1+k_2}|$ divides $|x^{k_1+k_2}|$. And since $G=langle x rangle$ and $H=langle y rangle$ with $|x|=m$ and $|y|=n$, it clearly follows that $G cong mathbb{Z}/mmathbb{Z}$ and $H cong mathbb{Z}/nmathbb{Z}$. Therefore
        $$|y^{k_1+k_2}|=frac{n}{gcd(k_1+k_2,n)},$$
        $$|x^{k_1+k_2}|=frac{m}{gcd(k_1+k_2,m)}.$$
        Using that the first expression divides the second, we have for $r in mathbb{Z}$
        $$|x^{k_1+k_2}|=frac{m}{gcd(k_1+k_2,m)}=|y^{k_1+k_2}|cdot r = frac{nr}{gcd(k_1+k_2,n)}$$
        Providing us with
        $$m=frac{gcd(k_1+k_2,m)}{gcd(k_1+k_2,n)} cdot nr$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 20 at 3:23









        Andrew TawfeekAndrew Tawfeek

        1,8031722




        1,8031722






























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