I'm trying to do 3.16 in Fraleigh's algebra book. Here it is:

The map $f: Z\to Z$ defined by$ f(n) = n + 1$ for $n$ in $Z$ is 1-1 and onto $Z$. Give the definition of a binary operation $\ast$ on $Z$ such that $f$ is an isomorphism mapping $\langle Z, +\rangle$ onto $\langle Z, \ast\rangle$.

I know isomorphisms have to pass the homomorphism property:

$$f(mn) = f(m)f(n)$$

So:$ (mn) + 1 = (m + 1)(n + 1) = nm + n + m + 1$
So $0 = n + m$...

A bit lost at this point.

Any guidance would be great!

Thanks, Mariogs


You need to define the operation $*$ on $\mathbb Z$ in order to make $f$ and isomorphism.

So we need for $f$ to satisfy the homomorphism property: $$f(m+n) = m+n+1 = f(m)*f(n) = (m+1)*(n+1)$$

Now, what operation $*$ will give us the equality: $$m+n + 1 = (m+1)*(n+1)\;\;?$$

How about: $p*q = p + q -1$. Compute, now, $(m+1)*(n+1)$, and see if you obtain the desired $m+n +1$.

  • $\begingroup$ ahhh, that makes a lot more sense! $\endgroup$ – bclayman Feb 19 '14 at 16:20
  • $\begingroup$ You're welcome! Feel free to post your definition of $*$. $\endgroup$ – Namaste Feb 19 '14 at 16:30
  • 1
    $\begingroup$ If it helps mentally, BTW, since $*$ so strongly evokes multiplication, you could (at least in your own work) write the operation as $a\Diamond b$ and talk about an isomorphism from $(\mathbb{Z}, +)\to(\mathbb{Z}, {}\Diamond{})$. $\endgroup$ – Steven Stadnicki Feb 19 '14 at 16:36

From your original question and also your comments, I think you are a little confused about what it means for $f$ to be a homomorphism. You said that $f$ is a homomorphism if $$f(mn) = f(m)f(n).$$ But this is wrong. That is what it means for $f$ to be a homomorphism from $\langle Z, \cdot\rangle$ to $\langle Z, \cdot\rangle$.

But here we want to define a new operation, called $\ast$, so that $f$ is a homomorphism from $\langle Z, +\rangle$ to $\langle Z, \ast\rangle$.

The property for that to be true is that $$f(m+n) = f(m)\ast f(n).$$

Notice that there is no multiplication anywhere in this formula. There is addition, and there is $\ast$, which is not multiplication.

You know $f$, so you know everything about what this formula says except what $\ast$ means. You can use algebra to figure out what $a\ast b$ must be in order for the formula to hold.


Hint: Multiplication isn't involved in the problem. The homomorphism property for $f:(\mathbb Z,+)\to(\mathbb Z,\ast)$ states that $f(m+n)=f(m)\ast f(n)$. Now apply the definition of $f$ and you should be back on track.

  • $\begingroup$ ah so this should really be mn + 1 = (m+1) * (n+1), where * is addition mod 8. So I get mn + 1 = m + n + 2, so mn = m + n + 1...and then? $\endgroup$ – bclayman Feb 19 '14 at 16:14
  • $\begingroup$ I'm not sure where addition mod 8 came from. If * was defined as addition mod 8 somewhere else in the textbook, then it doesn't have the same meaning here. In abstract algebra, you end up studying lots of different operations, and symbols will inevitably be reused. Anyway, it sounds like the other answers helped! $\endgroup$ – Chris Culter Feb 19 '14 at 16:52

Define $\ast$ as $a \ast b = a+b-1$.

We then need to check $$\phi(a+b)=\phi(a)\ast\phi(b).$$

By definition, $$\phi(a+b)=a+b+1.$$

Also, by definition $$\phi(a)\ast \phi(b)=(a+1)\ast(b+1)=(a+1)+(b+1)-1=a+b+1.$$

Thus $$\phi(a+b)=\phi(a)\ast \phi(b),$$ and $\phi$ is an isomorphism.

  • $\begingroup$ how would you get to that by making use of the homomorphism property though? $\endgroup$ – bclayman Feb 19 '14 at 16:19
  • $\begingroup$ @Marilogs I've re-edited my answer to provide more detail, hope this helps. Note, you seem to get confused between an abstract binary operation, $\ast$, on pairs of elements in a group and standard operations, e.g. addition or multiplication. $\endgroup$ – David Simmons Feb 19 '14 at 16:30

Their is a general procedure for transferring structures over bijections of sets. What do I mean by this? Well consider the following proposition:

Proposition Let $(G,\bullet)$ be a group and $S$ be a set, such that their is a bijection of sets, $$\phi:G\to S$$ (note that $\phi$ is not a group homomrphism, yet). (Note also that a in category theory you would say that their is a set isomorphism $\phi:UG\to S$, where $U$ is the forgetful functor to sets.) Then their is a unique group operation, $\star$ on $S$ that makes $\phi$ a group homomorphism.

sketch of proof: Lettuce start by defining the group operation $\star$. Define: $$\textbf{1}:s_1\star s_2=\phi(\phi^{-1}(s_1)\bullet\phi^{-1}(s_2))$$

We will now show that this product is associative.

$$(s_1\star s_2)\star s_3=\\ \phi[\phi^{-1}(s_1\star s_2)\bullet \phi^{-1}(s_3)]=\\ \phi[\phi^{-1}(\phi[\phi^{-1}(s_1)\bullet \phi^{-1}(s_2)])\bullet \phi^{-1}(s_3)]=\\ \phi[\phi^{-1}(s_1)\bullet [\phi^{-1}(s_2)\bullet \phi^{-1}(s_3)]]=\\ \phi[\phi^{-1}(s_1)\bullet \phi^{-1}(\phi([\phi^{-1}(s_2)\bullet \phi^{-1}(s_3)]])=\\ \phi(\phi^{-1}(s_1)\bullet\phi^{-1}(s_2\star s_3))=\\ s_1\star(s_2\star s_3).$$

I will leave the other axioms to you. $$\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\Box$$

Now what use is this proposition? Well in you situation, you have the following: A group, $(\mathbb{Z},+)$ and you are looking for another group operation on $\mathbb{Z}$ so that the bijection, $f(n)=n+1$ is a homomorphism. The second $\mathbb{Z}$ is just a set, without any group structure. So let us define $$n_1*n_2=f(f^{-1}(n_1)+f^{-1}(n_2)).$$ This is simply applying $\textbf{1}$ to our situation. Let us now work out what this must be.


You might say that we had to work hard to get this. But this proposition is quite powerful. What the answer does is to solve a more general problem. A few exercises that you could try (if you want) are as follows:

  1. Define a group operation, $*$ on $\mathbb{Z}$ such that $f(n)=n+5$ is a homomorphism from $(\mathbb{Z},+)$ to $(\mathbb{Z}, *)$ using this proposition.

  2. Define a group operation $*$ on $\mathbb{Z}/2\oplus\mathbb{Z}/4$ such that $f(m,n)=(m,n+1)$ is a homomorphsm from $(\mathbb{Z}/2\oplus\mathbb{Z}/4,+)$ to $(\mathbb{Z}/2\oplus\mathbb{Z}/4,*)$

  3. Consider the map, $f:\mathbb{Z}/3\to\{a,b,c\}$, such that $f(0)=a, f(1)=b, f(2)=c$. Use this proposition to to define a group operation $*$ on $\{a,b,c\}$ such that $f$ is a group homomorphism from $(\mathbb{Z}/3,+)$ to $(\{a,b,c\},*)$.

  4. (this one is harder) If you have a ring, $R$ and a set $S$, and a set bijection $f:R\to S$ Define a ring structure on $S$ such that $f$ is a ring homomorphism.

Note also Propositions like the above are often called "transfer of structure theorems". So is the proposition indicated in exercise 4. I learned this proposition from Seth Warner's book "Modern Algebra". This is Theorem 6.3 on page 43 (of the copy of the book that I have). He calls this theorem the "Transplanting Theorem".


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