# How can we show that $\langle u,v^2\rangle \cong \mathbb Z_2\times \mathbb Z_4$, where $u,v^2\in\langle u,v :u^2=v^8=1,vu=uv^5\rangle$?

Let $M=\langle u,v : u^2=v^8=1, vu=uv^5\rangle$ be the modular group of order $16$. Prove that $\langle u, v^2\rangle\cong\mathbb Z_2\times\mathbb Z_4$.

One of the most elementary ways to show that two groups are isomorphic is to find a bijective homomorphism involving the two groups. I'm tempted to say we define the map $$\varphi:\langle u, v^2\rangle\to\mathbb Z_2\times\mathbb Z_4, \;\; \varphi(u) = (\overline 1, \overline 0), \varphi(v^2) = (\overline 0, \overline 1).$$ Then it suffices to check that $(\overline0,\overline 1)$ and $(\overline 1, \overline 0)$ satisfy the same properties from $M$. But I unfortunately got slightly confused while trying to prove this. Here's what I had: $$(\overline1,\overline0)^2=(\overline1+\overline1,\overline0+\overline0) = (\overline0,\overline0)$$ $$(\overline0,\overline1)^4 = (\overline{0+0+0+0},\overline{1+1+1+1}) = (\overline0,\overline0)$$ \begin{align}vu &= uv^5\\ \implies v &= uv^5u^{-1} \\ \implies v^2 &= uv^5u^{-1}uv^5u^{-1}\\ &= uv^{10}u^{-1}\\ &= uv^2u^{-1}. \tag{*}\end{align} So, we have $$(\overline1,\overline0)(\overline0,\overline1)(\overline1,\overline0) = (\overline{1+0+1},\overline{0+1+0}) = (\overline0,\overline1),$$ which shows that $*$ is satisfied. Hence there does exist a homomorphism $\varphi:\langle u,v^2\rangle \to \mathbb Z_2\times\mathbb Z_4$ such that $\varphi(u)=(\overline1,\overline0)$ and $\varphi(v^2) = (\overline0,\overline1)$. (!!) And it is extremely clear that $(\overline0,\overline1)$ and $(\overline1,\overline0)$ generate $\mathbb Z_2\times\mathbb Z_4$. Thus $\varphi$ is a surjection, and $|\langle u,v^2\rangle| = 8 = |\mathbb Z_2\times\mathbb Z_4|$ (given). Therefore it must be that $\langle u,v^2\rangle\cong\mathbb Z_2\times\mathbb Z_4$.

A question I had: Is (!!) a valid conclusion to draw? They seem to draw this conclusion in Dummit and Foote on page 39, but I'm not entirely convinced that we can draw this conclusion. Thanks for your help.

• I am "sort of" convinced. I think it would be clearer if you replaced "suffices to check that $(\bar{0},\bar{1})$ and $(\bar{1},\bar{0})$ satisfy the same properties from $M$" by "suffices to check that $(\bar{0},\bar{1})$ and $(\bar{1},\bar{0})$ satisfy the same properties in $\mathbb{Z}_2 \times \mathbb{Z}_4$ as $u$, $v^2$ do in $M$." – ancientmathematician Mar 28 '17 at 6:47

The group $\langle u, v^2\rangle$ is a subgroup of $M$. Now the relators of this group are $[u^2 = 1, (v^2)^4 = 1, uv^2u = uvuuvu = v^5v^5=v^{10} = v^2]$, so $u$ and $v^2$ commute and so generate $\Bbb Z_2\times\Bbb Z_4$.