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$G$ is generated by $a$ and $b$ satisfying the relators $a^4$, $b^2$, $(ab)^2$.

Let $a = (123)$ and $b = (12)$. It can be solved that $a^4=e$, $b^2=e$, and $(ab)^2=e$. So there is an epimorphism phi that maps $G$ to $D_4$ (Van Dyck's Theorem). What is this order of G is greater or equal to order of D4?

Can anyone help on the procedure, the next step l think is to show that phi is one to one so that $G$ is isomorphic to $D_4$? Thank you.

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  • $\begingroup$ Hint: the relator $(ab)^2$ allows you to put the elements of the group in the form $a^ib^j$. Use this to give an upper bound on the order of the group. $\endgroup$
    – user1729
    Commented Sep 11, 2017 at 7:45
  • $\begingroup$ Perhaps that you meant to write $a=(1234)$. After all, $(123)^4=(123)$. $\endgroup$
    – José Carlos Santos
    Commented Sep 11, 2017 at 8:04
  • $\begingroup$ Is b = (12) right? $\endgroup$
    – Romeo
    Commented Sep 11, 2017 at 8:22
  • $\begingroup$ @RomeoAlma No, $\langle (1234),(12)\rangle=S_4$. $\endgroup$
    – Dietrich Burde
    Commented Sep 11, 2017 at 12:32

1 Answer 1

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This is Theorem $1.1$ in K. Conrad's notes, which are a very good reference for dihedral groups. The proof there is for all $D_n$, $n\ge 3$, and is very detailed. It is shown that $$\langle a,b\rangle=\{e,a,a^2,\ldots ,a^{n-1},a^2b,a^3b,\ldots,a^{n-1}b\}$$ has $2n$ elements, and an isomorphism to $D_n$ is constructed.

One may realize $D_4$ as a subgroup of $S_4$ in various ways, e.g., by $$ a=(1234),\; b=(14)(23), $$ see this MSE-question.

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