Mdthbs

I was trying to define a homomorphism between some finite groups and I had the following idea.

Suppose that $G = langle a, brangle $, and $H = langle x, yrangle$. We define $varphi:G to H$ by $varphi(a)=x,$ $varphi(b)=y$, and for an arbitrary element of $G$, $g_1^{epsilon_1}g_2^{epsilon_2} cdots g_n^{epsilon_n}$ (where each $g_i$ is either $a$ or $b$, and each $epsilon_i$ is either $1$ or $-1$), then $varphi(g_1^{epsilon_1}g_2^{epsilon_2} cdots g_n^{epsilon_n}) = varphi(g_1)^{epsilon_1} varphi(g_2)^{epsilon_2} cdots varphi(g_n)^{epsilon_n}$.

If this is well-defined, then this is obviously a homomorphism. However, I do not know if this is well defined. I suspect not.

Any arbitrary map of generators to generators need not extend to a homomorphism, even if there is only one generator. For example, there is no nontrivial homomorphism between two cyclic groups of different prime orders.

The problem is that any combination of generators that results in the identity in the original group must also evaluate to the identity when you map to the target generators. Otherwise the map would not be a homomorphism.

Look at it this way. If what you propose was well-defined in general, then your $varphi$ would always have inverse $psicolon Hto G$ just by setting $psi(x)=a$, $psi(y) = b$. But not all groups generated by two elements are isomorphic. For example look at $mathbb Z_2timesmathbb Z_2$, $mathbb Ztimes mathbb Z_2$, $mathbb Ztimesmathbb Z$, $F_2$ (free group generated by two elements), and any dihedral group $D_n$.

In general, for the group $G$ to be generated by two elements, it means that there is an epimorphism $varepsiloncolon F_2to G$ and by the first isomorphism theorem $Gcong F_2/kervarepsilon$. So, to define $varphicolon Gto H$, you need to have a homomorhpism $psicolon F_2 to H$ such that $kervarepsilonsubseteq kerpsi$, by the fundamental homomorphism theorem.

To explain what it means precisely, let $F_2 = langle x,yrangle$ and $G = langle a,brangle$. In this case, $varepsilon$ is just $xmapsto a$ and $ymapsto b$. To specify $psi$, it is enough to pick any two elements $h_1,h_2in H$ and let $psi(x) = h_1$, $psi(y) = h_2$. This always does define homomorphism - this is by definition of a free group, and this is essentially what you wanted to do, but for group $G$. Now, condition $kervarepsilonsubseteq kerpsi$ is equivalent to saying that for all $a^{alpha_1}b^{beta_1}ldots a^{alpha_n}b^{beta_n} = e_G$, it must be that $h_1^{alpha_1}h_2^{beta_1}ldots h_1^{alpha_n}h_2^{beta_n} = e_H.$ If that's true, you can set $varphi(a) = h_1$, $varphi(b) = h_2$, and it will be well-defined.

For more concrete example, let's take $G = mathbb Z_2timesmathbb Z_2$ and $H = mathbb Ztimesmathbb Z$. Let $a = (1,0)$ and $b = (0,1)$. Note that $2a = 2b = (0,0)$ (I've switched to additive notation, as is customary). However, for any hypothetical $varphicolon Gto H$, $2varphi(a) = varphi(2a) = varphi(0,0) = (0,0)$. Since the only $hin H$ such that $2h = (0,0)$ is $h = (0,0)$, we need to have $varphi(a) = (0,0)$ and similarly $varphi(b) = (0,0)$. So, the only homomorphism is trivial.

I think it is not, in general, well-defined. If it is indeed a well-defined map, then it is a surjection from $G$ onto $H$. Also, we can similarly define a surjection of $H$ onto $G$. But then we would have that any two groups generated by $2$ elements have the same cardinality, which is probably not true.