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Given two groups $(G_A, *_A)$ and $(G_B, *_B)$, can we define a group isomorphism $f : G_A \rightarrow G_B$ by

  1. $G_A$ and $G_B$ have the same number of elements
  2. with matrix $A$ such that $A_{ij}=i*_Aj$ and matrix $B$ such that $B_{ij}=i*_Bj$, then for all $i,j,x,y \in G_A$, if $A_{ij}=A_{xy}$ then $B_{f(i)f(j)}=B_{f(x)f(y)}$

and if so, how to prove that this definition is equivalent to the standard one?

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Confusing to use $A$ and $B$ as matrices and also as subscripts that distinguish the groups. – Gerry Myerson Sep 29 '12 at 6:06
@GerryMyerson why is that? – idonutunderstand Sep 29 '12 at 23:06
up vote 7 down vote accepted

I don't think this is equivalent. For if, for example, $G_A = G_B = \mathbb{Z}/2\mathbb{Z}$ (thought of as the set $\{0,1\}$), but we define $f(0) = 1$ and $f(1) = 0$, then I think $f$ is an isomorphism according to your definition, but it isn't actually an isomorphism.

First, it's not an isomorphism (according to the usual notion of isomorphism) because it doesn't map the identity to the identity, so it's not even a homomorphism.

On the other hand, $$A = B = \begin{bmatrix} 0 & 1\\1&0 \end{bmatrix}.$$

Then, one can check that $A_{00} = A_{11}$ and $B_{f(0)f(0)} = B_{f(1)f(1)}$, and similarly that $A_{01} = A_{10}$ and $B_{f(0)f(1)} = B_{f(1)f(0)}$.


In fact, it's worse than I thought. In the example I gave, the 2 groups were actually isomorphic, but we have the following fact:

If $G_A$ and $G_B$ have the same number of elements, then they must be "isomorphic" according to your definition. Let $f:G_A\rightarrow G_B$ be any constant map. Then we have $B_{f(a)f(b)} = B_{f(x)f(y)}$ for any choices of $a,b,x,y$.

So, for example, $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}2\mathbb{Z}$ and $\mathbb{Z}/4\mathbb{Z}$ are "isomorphic" according to your definition, but they aren't isomorphic at all.

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The condition (which I’ll call $P$) actually does say something interesting if we add the assumption that $f$ is a bijection.

Theorem. Let $G$ and $H$ be groups. A bijection $f:G\to H$ has property $P$ iff there are an isomorphism $h:G\to H$ and a $z\in Z(H)$ such that $f(g)=h(g)z$ for each $g\in G$.

Proof. Suppose first that $f$ has property $P$. Note first that if $a,b\in G$ and $ab=ba$, then $f(a)f(b)=f(b)f(a)$, so $f(1_G)\in Z(H)$. Let $z=f(1_G)$, and define $h:G\to H:g\mapsto f(g)z^{-1}$. Then for $a,b\in G$ we have $$h(a)h(b)=f(a)z^{-1}f(b)z^{-1}=f(a)f(b)z^{-2}=f(ab)f(1_G)z^{-2}=f(ab)z^{-1}=h(ab)\;,$$

since $ab=(ab)1_G$, so $h$ is an isomorphism.

Conversely, if $h:G\to H$ is an isomorphism, and $z\in Z(H)$, let $f:G\to H:g\mapsto h(g)z$. If $a,b,c,d\in G$ and $ab=cd$, then

$$f(a)f(b)=h(a)zh(b)z=h(ab)z^2=h(cd)z^2=h(c)zh(d)z=f(c)f(d)\;,$$ and $f$ has property $P$. $\dashv$

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