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$\begingroup$

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 $a^{2k}b^{2k} \equiv 1 \mod n$.
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    $\begingroup$ Hint: in a group, if $g^k=e$ then the order of $g$ divides $k$. $\endgroup$
    – lulu
    Commented Sep 22, 2023 at 4:48
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    $\begingroup$ @ronno Oh, no. Suppose $b=a^{-1}$. $\endgroup$
    – lulu
    Commented Sep 22, 2023 at 4:57
  • $\begingroup$ @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. $\endgroup$ Commented Sep 22, 2023 at 6:07
  • $\begingroup$ 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).\ \ $ $\endgroup$ Commented Sep 22, 2023 at 7:05
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    $\begingroup$ 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. $\endgroup$
    – ronno
    Commented Sep 22, 2023 at 9:36

1 Answer 1

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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.

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    $\begingroup$ should it be $o(a.b) | l.c.m.(o(a),o(b))$ rather than =? $\endgroup$
    – Mzq
    Commented Sep 23, 2023 at 2:10
  • $\begingroup$ @Mzq yes you are right i correct it now. $o(a.b)=l.c.m.(o(a),o(b))$ if G is abelian and $\langle a\rangle \cap \langle b \rangle={e}$ $\endgroup$ Commented Sep 23, 2023 at 6:23
  • $\begingroup$ do I need to prove that G is abelian? or I can just use it $\endgroup$
    – Mzq
    Commented Sep 25, 2023 at 12:48
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    $\begingroup$ @Mzq in fact $G\leq$ $\mathbb R$ and $\mathbb R $ with the usual multiplication is abelian so any subgroup will be abelian $\endgroup$ Commented Sep 25, 2023 at 13:10

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