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I was reading this page Probability that cubic of integer ends with 11 and one of the answers states:

"If $^3 ≡ 1$ mod$ 100$, then $^2 ≡ 1$ and so is a unit mod $100$. The group of units mod $100$ has order $\phi(100)=40$. Therefore, since gcd$(40,3)=1$, the map $ \mapsto ^3$ is a bijection. Thus, there is exactly one such that $^3≡11$ mod $100$."

My question is why the gcd$(40,3)=1$ tells us the map is bijective. I see the $3$ comes from our map being $x$ to $x^3$, but don't see why this forces it to be a bijection. (I see we can just bash this out, so the statement is indeed correct).

BTW: I didn't leave a comment on the initial page because I don't have the ability to do so.

Bill Dubuque
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Sebastian1213
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  • Hint: argue that $x^3=y^3$ would imply the existence of an element of order $3$. – lulu Jul 26 '24 at 10:29
  • Use Bézout to prove that the map is subjective. – lhf Jul 26 '24 at 11:12
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    From $\gcd(40,3)=1$, it follows that $3$ is invertible mod $40$, so there exists $a$ such that $3a\equiv1\bmod40$, so $x\mapsto x^a$ is an inverse of $x\mapsto x^3$ for units $x$ mod $100$ – J. W. Tanner Jul 26 '24 at 12:29

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The group of units of $\mathbb{Z}/100\mathbb{Z}$ (its multiplicative group, if you like) has order $\phi(100)=\phi(2^2)\phi(5^2)=2\cdot 5\cdot 4=40$. If $x^3=y^3$ then $x=y$ or $xy^{-1}$ has order $3$. Suppose $x\ne y$ so $xy^{-1}$ has order $3$. The order of an element must divide the order of the group, so $3|40$ but this is a contradiction as $\operatorname{gcd}(3,40)=1$. So, the map $x\mapsto x^3$ is injective. Since we are talking about finite sets, an injective map $(\mathbb{Z}/100\mathbb{Z})^\times$ to itself must be bijective, so it is an automorphism.

Joshua Tilley
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  • Oh I see. Thank you! So the initial poster implied that since the gcd is one, there is no element of order three which we would need if the map was not a bijection. Is this step required in the logic and the original poster skipped it due to triviality, or is there another way to think about it? – Sebastian1213 Jul 26 '24 at 10:36
  • Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. – Bill Dubuque Jul 26 '24 at 19:06