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Suppose $A = \langle x\mid x^m = e\rangle$, $B = \langle y\mid y^n = e\rangle$, and $C = \langle z\mid z^d = e\rangle$ (these are cyclic groups of order $m,n,d$, respectively). Also suppose $d$ divides $m$ and $n$.

(a) Show that $\phi : C \rightarrow A \times B $, $ z \mapsto (x^{m/d}, y^{n/d})$ is an injective group homomorphism.

(b) Find the elementary divisors of $D = (A \times B)/\phi(C)$ and confirm that $D$ is cyclic iff $d=\mathrm{gcd}(m,n)$.

I think that this problem is related to the Chinese Remainder Theorem and its proof, but I am not sure how to proceed with this. I would appreciate any help with this. Thank you.

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What is $x$ and $y$ in your definition of $\phi$? –  Phira Nov 21 '12 at 20:32
    
There seems to be an error in your definition of $\phi$. If $z \mapsto x^{m/d}$, that just means $z \mapsto e^{1/d}$ (which is once again $e$, I think) and likewise for $y^{n/d}$: $y^{n/d}=e^{1/d}=e$. I am thinking your notation means $z \mapsto (x^{m/d}, y^{n/d})$, but I don't see how this is a mapping given that it is not a function in terms of some variable related to $z$. Is this ($\phi$) intentionally a constant function? –  000 Nov 21 '12 at 20:36
    
Is the purpose of this problem to show that $(x^{m/d},y^{n/d})$ reduces to $(e,e)$ given that $x^m=e$ and $y^n=e$? And hence, $\phi$ is an injective homomorphism? –  000 Nov 21 '12 at 20:44
    
@limitless $x^{m/d}$ is not $e$ unless $d=1$. –  Hagen von Eitzen Nov 21 '12 at 21:35
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@Limitless We are given the fact that $d$ is a divisor of $m$, i.e. $m=dk$ for some $k$. Now $x^{m/d}$ is used (maybe somewhat ambiguously) to denote $x^k$. Since $k$ is not a multiple of $m$, $x^k$ is not $e$. –  Hagen von Eitzen Nov 21 '12 at 22:07
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up vote 2 down vote accepted

(a) First you have to show that this defines a homomorphism, i.e that $Phi$ respects the relation $z^d=e$ in the presentation of $C$. Indeed, we have $$(x^{m/d},y^{n/d})^d = (x^{(m/d)d},y^{(n/d)d})=(x^m,y^n)=(e,e),$$ which is th eneutral element of $A\times B$. To show injectivity, assume $z^k\mapsto (e,e)$. That implies $x^{(m/d)k}=e$ and $y^{(n/d)}k=e$ and thus that $\frac md k$ is a multiple of $m$ and $\frac nd k$ a multiple of $n$. Both simply mean that $\frac kd$ is an integer, i.e., $k$ is a multiple of $d$.

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