1. ## Transitive

Prove that the notion of group isomorphism is transitive.

My proof so far:

Suppose G, H, and K are groups.

Let the function f be the isomorphism from G onto H, then $\displaystyle f(g_{1}g_{2})=f(h_{1})f(h_{2})$ with f is onto and 1-1.

Now let the function t be the isomorphism from H onto K, then $\displaystyle t(h_{1}h_{2})=t(k_{1})t(k_{2})$ with t is onto and 1-1.

My plan is to get a function, say X, such that $\displaystyle X(g_{1}g_{2}) = X(k_{1}k_{2})$

First I tried with X = f(t), but it doesn't really work out, so am I at least doing something right here?

Thanks.

Prove that the notion of group isomorphism is transitive.

My proof so far:

Suppose G, H, and K are groups.

Let the function f be the isomorphism from G onto H, then $\displaystyle f(g_{1}g_{2})=f(h_{1})f(h_{2})$ with f is onto and 1-1.
That equation should be $\displaystyle f(g_{1}g_{2})=f(g_{1})f(g_{2})$.

Now let the function t be the isomorphism from H onto K, then $\displaystyle t(h_{1}h_{2})=t(k_{1})t(k_{2})$ with t is onto and 1-1.
That equation should be $\displaystyle t(h_{1}h_{2})=t(h_{1})t(h_{2})$.

My plan is to get a function, say X, such that $\displaystyle X(g_{1}g_{2}) = X(k_{1}k_{2})$
That equation should be $\displaystyle X(g_{1}g_{2}) = X(g_{1})X(g_{2})$.

First I tried with X = f(t), but it doesn't really work out, so am I at least doing something right here?
The reason it doesn't "really work out" may be that you are trying to combine f and t in the wrong order. If you define $\displaystyle X(g) = t(f(g))$ then $\displaystyle X(g_1g_2) = t(f(g_1g_2)) = t(f(g_1)f(g_2)) = t(f(g_1))t(f(g_2)) = X(g_1)X(g_2)$. Of course, you also have to verify that X is onto and 1-1.

Prove that the notion of group isomorphism is transitive.

My proof so far:

Suppose G, H, and K are groups.

Let the function f be the isomorphism from G onto H, then $\displaystyle f(g_{1}g_{2})=f(h_{1})f(h_{2})$ with f is onto and 1-1.

Now let the function t be the isomorphism from H onto K, then $\displaystyle t(h_{1}h_{2})=t(k_{1})t(k_{2})$ with t is onto and 1-1.

My plan is to get a function, say X, such that $\displaystyle X(g_{1}g_{2}) = X(k_{1}k_{2})$

First I tried with X = f(t), but it doesn't really work out, so am I at least doing something right here?

Thanks.
If $\displaystyle \phi: G_1\mapsto G_2$ is an isomorphism and $\displaystyle \psi: G_2\mapsto G_3$ is an isomorphism then $\displaystyle \psi \circ \phi:G_1\mapsto G_3$ is an isomorphism.