1
$\begingroup$

If I want to show that two cyclic groups are isomorphic, is it enough to show that their cardinality is the same and that the generators of the groups are mapped onto each other?

To be precise: I am given a group G with

   G = <a,b> , |G| = 2q

and I am supposed to show that G is isomorphic to Dq, the Dihedral Group with

    Dq = <s,o> , |Dq| = 2q

is it enough to show that

    a --> s and b --> o

?

Thanks already for any help!

$\endgroup$
  • $\begingroup$ The dihedral groups $D_q$ aren't cyclic. Did you mean to say "finite" rather than "cyclic"? $\endgroup$ – bradhd May 3 '14 at 11:57
  • $\begingroup$ You're right. Sorry, about that. Neither G nor Dq are cyclic, so it shouldn't interfere with the isomorphism. $\endgroup$ – eva May 3 '14 at 12:32
1
$\begingroup$

If $G$ and $H$ are finite groups of the same cardinality, then any surjective homomorphism $\phi:G\to H$ is bijective, hence an isomorphism. If $H = \langle h_1,\ldots,h_n\rangle$ and each $h_i$ is in the image of $\phi$, then $\phi$ is surjective.

So in your situation, if you can prove that there is a homomorphism $\phi:G\to D_q$ such that $\phi(a)=s$ and $\phi(b)=o$, then indeed $\phi$ is an isomorphism. The question is whether there exists such a homomorphism. There does exist such a homomorphism if and only if $s$ and $o$ satisfy all the same relations together that $a$ and $b$ satisfy. This can be checked using a presentation of $G$.

A presentation of a group $G$ is a set $S$ of generators for $G$, together with a minimal set $R$ of relations on that set of generators. This means that every relation satisfied by the elements of $S$ can be derived from the relations in $R$. We usually write this as $G = \langle S | R \rangle$.

For example, one presentation of the dihedral group of order $2q$ is

$$D_q = \langle \rho,\tau | \rho^q=1, \tau^2=1, (\rho\tau)^2=1\rangle$$ where $\rho$ is a rotation and $\tau$ is a reflection. ($1$ is the identity element.)

So if $\phi:D_q\to H$ is a homomorphism, then we must have $$\phi(\rho)^q=1, \phi(\tau)^2=1, (\phi(\rho)\phi(\tau))^2=1.$$ The usefulness of a presentation is that these relations are equivalent to $\phi$ defining a homomorphism. That is, if $x$ and $y$ are elements of $H$ such that $$x^q=1, y^2=1, (xy)^2=1$$ then there exists a unique homomorphism $\phi:D_q\to H$ such that $\phi(\rho)=x$ and $\phi(\tau)=y$. The analogous statement is true for an arbitrary presentation $\langle S|R\rangle$.

$\endgroup$
  • $\begingroup$ Thanks for the answer! What exactly do you mean by "presentation" of G? I am not a native english speaker and I also don't study math in englisch, which is why I am not familiar with all of the mathematical vocabulary.. Are you maybe refering to Cayley's theorem that states that if G is finit (which it is in my case) then G is isomorphic to a subgroup of Sn (the symmetric group of degree n). Since Dq is also a subgroup of Sn s, o, a and b satisfy the same relations... $\endgroup$ – eva May 3 '14 at 12:37
  • $\begingroup$ I have edited my answer with more explanation. $\endgroup$ – bradhd May 3 '14 at 13:26
  • $\begingroup$ Awesome - thank you so much! I actually had to prove earlier on in the question that ord(a) = ord(b) = 2 and since G = <a,b> also ord(ab) = 2. This shows that s, o, a and b satisfy the same relations. Your answer was very helpful! $\endgroup$ – eva May 3 '14 at 18:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.