Groups, Subgroups, Lagrange's Theorem. What am I doing wrong?

Question

Let's take the group $(\mathbb{Z}_{10},\cdot)$ as an example, where $\mathbb{Z}_{10}$ contains the congruency classes modulo 10. Now let's take all the invertible elements of that group (all the elements that have an inverse $x^{-1}$ such that $xx^{-1}=[1]$) and call it $\mathbb{Z}_{10}^*$. If I'm not mistaken $\mathbb{Z}_{10}^* = \{[1],[3],[7],[9]\}$.

Now, it seems to me that $\mathbb{Z}_{10}^*$ is a subset of $\mathbb{Z}_{10}$. However, I know that for Lagrange's theorem if we have a subgroup $H$ of a group $G$ then $|H|$ divides $|G|$.

But in my example $|\mathbb{Z}_{10}^*| = 4$ and 4 does not divide $|\mathbb{Z}_{10}| = 10$.

I bet I'm wrong and not Lagrange, where did I fumble?

Answer 1

What am I doing wrong?

$\mathbb Z_{10}$ is not a group under multiplication; in a group, each element must have an inverse.

Answer 2

First of all, $(\Bbb Z_{10},\cdot)$ is not a group. Each element of a group must have an inverse; $[0]_{10}$ has no inverse with respect to multiplication modulo $10$.

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A subgroup $(H,\ast)$ of a group $(G,\star)$ must have the operation $\ast$ equal to the operation $\star$, restricted to the subset $H$ of $G$, such that:

• $\ast$ is closed on $H$.
• The identity $e_G$ of $G$ is in $H$.
• Each $h\in H$ has $h^{-1}\in H$ (with respect to $\ast$).

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Here $(\Bbb Z_{10}^*,\cdot)$ is a group under multiplication modulo $10$, not addition modulo $10$, so it does not have the same operation as $(\Bbb Z_{10},\color{red}{+_{10}})$ (which I assume is what you intended). Moreover, $[0]_{10}\notin\Bbb Z_{10}^*$, and yet $[0]_{10}$ is the identity of $(\Bbb Z_{10},+_{10}).$

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Therefore, we cannot apply Lagrange's Theorem since that only applies to subgroups.