# Preserving relations and operations in an automorphism

An automorphism is an isomorphism of an algebraic structure with itself.

Let's review an isomorphism $$i$$ of two linearly ordered groups $$G(+,<)$$ and $$F(*,<)$$.

$$i$$ is linear order-preserving when $$a < b \iff i(a) < i(b)$$ for any elements $$a$$ and $$b$$ in $$G$$.

Let's note that every linear order induces the reverse order: $$G(+, <, >)$$, $$F(*, <, >)$$.

An isomorphism $$i^*$$ is linear order-reversing when $$a < b \iff i(a) > i(b) \iff i(b) < i(a)$$.

Any linear order-preserving isomorphism $$i$$ induces linear order-reversing isomorphism $$i^*$$.

In case $$G$$ and $$F$$ are different, choosing between $$<$$ and $$>$$ in each of the groups is an arbitrary decision, so one can say $$i$$ and $$i^*$$ are equivalent.

It looks like this is not the case when we talk about automorphisms.

Let's assume $$G$$ is commutative and review the inversion automorphism $$i(a) = -a$$.

$$i$$ is order-reversing since $$a < b \iff -b < -a$$.

But in this case we can clearly see the difference between order-preserving and order-reversing automorphisms: $$i$$ maps $$<$$ to $$>$$ on the same set.

I am wondering if it make sense to explicitly say if an automorphism keeps operations and relations the same?

I've never thought of mapping $$+$$ to $$\times$$ in an automorphism of an algebraic structure $$(+, \times)$$.

Are there practical examples of such mapping?

There was a similar discussion with no positive conclusion in here: https://www.researchgate.net/post/Difference_between_an_order_automorphism_and_order_isomoprhism

• I don't do order theory, but I think that an "order-reversing" isomorphism $(G,<_G)\to (H,<_H)$ is better thought of as an isomorphism to the group $(H,>_H)$, or perhaps as an isomorphism from the domain $(G,>_G)$, much in the same way as we view a contravariant functor $F:C\to D$ as being actually a functor from $C^{\text{op}}$ to $D$. In this light, order reversing isomorphisms between two are not isomorphisms between the the initial objects but isomorphisms between different objects. Thus automorphisms of $(G,<_G)$ should be order-preserving under this viewpoint. – jgon Apr 4 at 21:44
• Please don't use $...$ for words in italics, such as $order-preserving$. The correct markup is *order-preserving*. Using $ for math symbols such as $i\$ is good, of course. – egreg Apr 4 at 22:04
• @egreg: Got it. Thank you. – Alex C Apr 4 at 23:06