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









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3080043%2fhomomorphism-xi-mapsto-yi-and-the-order-of-x-and-y%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    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









    $endgroup$


















      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






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3080043%2fhomomorphism-xi-mapsto-yi-and-the-order-of-x-and-y%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Mario Kart Wii

            Image
            Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
            Read more

            The Binding of Isaac: Rebirth/Afterbirth

            Image
            Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
            Read more

            What does “Dominus providebit” mean?

            Image
            5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
            Read more