Please Login to access more options.

Problem 68 (The Orders Of Elements Match In Isomorphic Groups)

Suppose that $f:G\to H$ a group isomorphism. Suppose that $f(g)=h$. Prove that the order of $g$ and the order of $h$ are the same, in other words show that $|g|=|h|$. In particular, since the only elements of order 1 in each group are the identities $e_G\in G$ and $e_H\in H$, then we must have $f(e_G)=e_H$.

Click here to see a partial proof that's missing one piece.

Why is the following proof not quite complete? It's missing something important. Please fill in the missing gap.

Suppose that $f:G\to H$ is an isomorphism and that $e_G$ and $e_H$ are the identities in $G$ and $H$ respectively. We first show that $f(e_G)=e_H$. Note that $e_Ge_G=e_G$. Apply $f$ to both sides to obtain $f(e_G)f(e_G)=f(e_G)$. Since $f(e_G)=f(e_G)e_H$, we now have $f(e_G)f(e_G)=f(e_G)\cdot e_H.$ Cancelling $f(e_G)$ from the left of both sides gives $f(e_G)=e_H$ as desired.

We now prove that if $a$ has order $n$, then $f(a)$ must have order $n$ as well. Suppose $a\in G$ has order $n$. This means that $n$ is the smallest positive integer such that $a^n=e_G$, or that $aa \cdots a=e$ (repeated $n$ times). If we apply $f$ to both sides of this equation, then we see that $f(a)f(a)\cdots f(a)=f(e_G)$. This means that $f(a)^n=f(e_G)$. Since we know $f(e_G)=e_H$, then we have shown $f(a)^n=e_H$ is the identity. This means the order of $f(a)$ must equal $n$. $\square$

What's missing? Pay close attention to the definition of order.

Solution

Let $G$ be a group. Let $H$ be a group such that $H$ is isomorphic to $G$. Let $f:G\to H$ be a group isomorphism.

We will begin by first giving an induction proof that will be used later. Let $S_n$ be the statement $f(g_1g_2\cdots g_n)=f(g_1)f(g_2)\cdots f(g_n)$ for all $g_1,g_2,\cdots,g_n\in G$. Because $f$ is a homomorphism, we know $S_2$ is true. Thus the base case is satisfied. Now suppose that for some $k$ we have that $S_k$ is true. Pick $g_1,g_2,\cdots,g_k,g_{k+1}\in G$. Let $s=g_1,g_2,\cdots,g_k$. Thus $s\in G$ because $G$ is a group. We compute $$\begin{align} f(g_1g_2\cdots g_kg_{k+1})&=f(sg_{k+1}) &\text{definition of }s \\ &= f(s)f(g_{k+1}) &f\text{ is a homomorphism} \\ &= f(g_1g_2\cdots g_k)f(g_{k+1}) &\text{definition of }s \\ &= f(g_1)f(g_2)\cdots f(g_k)f(g_{k+1}). &S_k\text{ is true} \end{align}$$

Thus we have $S_k$ implies $S_{k+1}$. Thus by mathematical induction, we know that $f(g_1g_2\cdots g_n)=f(g_1)f(g_2)\cdots f(g_n)$ is true for all $n\in\mathbb{N}$.

We will next show that the identity $e_G$ in $G$ maps to the identity $e_H$ in $H$. Pick $g\in G$ such that $g$ has finite order $n$. Let $h=f(g)$. Using identity properties we thus have $f(ge_G)=he_H$. Because $f$ is an isomorphism, we know $f(ge_G)=he_H$ implies $f(g)f(e_G)=he_H$. Recall $h=f(g)$. Thus $f(g)f(e_G)=he_H$ implies $hf(e_G)=he_H$. Lastly, given group cancellation laws, we have $f(e_G)=e_H$. Thus the identity $e_G$ in $G$ maps to the identity $e_H$ in $H$.

Using the information from above we compute $$\begin{align} e_H&=f(e_G) \\ &= f(g^n) \\ &= f(gg\cdots g) \\ &= f(g)f(g)\cdots f(g) \\ &= hh\cdots h \\ &= h^n. \end{align}$$ Therefore, we know $|h|\leq n$. Now let $k=|h|$. We compute $$\begin{align} e_H&=h^k \\ &=hh\cdots h \\ &= f(g)f(g)\cdots f(g) \\ &= f(gg\cdots g) \\ &= f(g^k). \end{align}$$ Since the identity in $G$ maps to the identity in $H$, we know that $e_H=f(g^k)$ implies $g^k=e_G$. Thus $|h|=k\geq n$. Thus because $|h|\leq k$ and $|h|\geq k$ we know $|h|=k$. Thus $|g|=|f(g)|$.

Tags
  • When you are ready to submit this written work for grading, add the phrase [[!Submit]] to your page. This will tell me that you have completed the page (it's past rough draft form, and you believe it is in final draft form). Don't type [[!Submit]] on a rough draft.
  • If I put [[!NeedsWork]] on your page, then your job is to review what I've written, address any comments made, and then delete all the comments I made. When you have finished reviewing your work, leave [[!NeedsWork]] on your page and type [[!Submit]]. (Both tags will show up). This tells me you have addressed the comments.
  • I'll mark your work with [[!Complete]] after you have made appropriate revisions.