# $G = \{c \in \{1, 2, \ldots, n-1\} \mid \gcd(c,n) = 1\},H = \{c \in G \mid \text{ord}(c) \text{ is odd}\}$ Proof H is closed under multiplication

Defining the Groups:

• $$G = \{c \in \{1, 2, \ldots, n-1\} \mid \gcd(c,n) = 1\}$$ represents the group of units mod $$n$$. These are the integers in the range $$1$$ to $$n-1$$ that are coprime to $$n$$.
• $$H = \{c \in G \mid \text{ord}(c) \text{ is odd}\}$$ represents the subset of elements in $$G$$ whose multiplicative order modulo $$n$$ is odd.

Proving that H is a Subgroup of G:

## Attempt

For H to be a subgroup of G, it must satisfy the following properties:

• H is non-empty.
• Closure: For any two elements x and y in H, the result of the operation x * y is also in H.
• Associativity: The operation * is associative on H.
• Identity Element: The identity element of G is also in H.
• Inverse Element: For each element x in H, the inverse x⁻¹ (in G) is also in H.

Proof:

• H is closed under inversion.
• Let $$k=ord(c)$$ it implise $$c^k\equiv 1 \mod n \to (c^k)^{-1}\equiv 1 \to (c^{-1})^{k}\equiv 1 \to c^{-1} \in H$$
• Identity element 1 exist.

The first try on $$H$$ is closed under multiplication:

• Suppose $$a$$ and $$b$$ are two elements in $$H$$. This means $$\text{ord}(a)$$ and $$\text{ord}(b)$$ are odd.
• Now, consider the element $$ab$$. We want to show that $$\text{ord}(ab)$$ is odd.
• Let $$\text{ord}(a)=2c+1$$, $$\text{ord}(b)=2d+1$$, WLOG let 2c+1 < 2d+1, then we have $$\mod n: a^{2c+1} \equiv 1, b^{2d+1} \equiv 1 \to a^{2c+1} b^{2d+1} \equiv 1 \to (ab)^{2c+1} b^{2(d-c)} \equiv 1$$.

The second try with proof by contradiction:

• If $$\text{ord}(ab)$$ were even, then we'd have $$(ab)^{2k} \equiv 1 \mod n$$, which implies $$a^{2k}b^{2k} \equiv 1 \mod n$$.
• Hint: in a group, if $g^k=e$ then the order of $g$ divides $k$.
– lulu
Commented Sep 22, 2023 at 4:48
• @ronno Oh, no. Suppose $b=a^{-1}$.
– lulu
Commented Sep 22, 2023 at 4:57
• @ronno yeah I think you’re thinking of direct products. If $(g_1,g_2)\in G_1\times G_2$, we have that $|(g_1,g_2)|=\operatorname{lcm}(|g_1|,|g_2|)$, and this can be extended to direct products of any finite length. Commented Sep 22, 2023 at 6:07
• Hint: odds are closed under products and divisors (i,e, they form a saturated monoid). The claim generalizes from odds to any such saturated monoid since $\,o(ab)\mid o(a)o(b).\ \$ Commented Sep 22, 2023 at 7:05
• Oops, I should have said that in an abelian group the order of $ab$ divides the lcm of the orders of $a$ and $b$, not is the lcm. Commented Sep 22, 2023 at 9:36

note that if G is abelian group then $$o(a.b)|l.c.m.(o(a),o(b))$$ . Let $$a,b\in H$$ then $$o(a)=2k+1$$ and $$o(b)=2k'+1$$ since $$l.c.m.$$ of 2 odd number is odd and any divisor of odd number is odd number , then since $$o(a.b)|l.c.m.(o(a),o(b))$$ we get $$o(a.b)$$ is odd therefore $$a.b\in H$$ so $$H$$ is closed under the multiplication.
• should it be $o(a.b) | l.c.m.(o(a),o(b))$ rather than =?
• @Mzq yes you are right i correct it now. $o(a.b)=l.c.m.(o(a),o(b))$ if G is abelian and $\langle a\rangle \cap \langle b \rangle={e}$ Commented Sep 23, 2023 at 6:23
• @Mzq in fact $G\leq$ $\mathbb R$ and $\mathbb R$ with the usual multiplication is abelian so any subgroup will be abelian Commented Sep 25, 2023 at 13:10