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Let $$G = \langle g \rangle$$ be a cyclic group of order $$n = 480 = 2^5 \cdot 3^1 \cdot 5^1$$, $H$ is a subgroup of $G$, and $m$ is the smallest natural number for which $g^m \in H$. Determine for which natural numbers $m$ the following conditions are satisfied:

(a) The index $$|G : H|$$ of $G$ by $H$ is divisible by $2$.

(b) The order $$|H|$$ of $H$ is divisible by $2$.

(c) The index $$|G : H|$$ of $G$ by $H$ is divisible by $2$ and not divisible by $3$.

(d) The order $$|H|$$ of $H$ is divisible by $2$ and not divisible by $3$.

Since $G$ is a cyclic group and $H$ is a subgroup of $G$, $H$ is also cyclic. Then, because $g^m$ is from $H$ and $m$ is the smallest such natural number, $H = \langle g^m \rangle$, and $|H| = |\langle g^m \rangle| = \frac{n}{\gcd(n,m)}$. By Lagrange's theorem, we then have $|G| = |H| \cdot |G:H| = \frac{n}{\gcd(n,m)} \cdot |G:H| \Rightarrow |G:H|=\gcd(n,m)$.

For (A), we have that $\gcd(n,m)$ should be divisible by $2$, that means that the prime factorization of $m$ should look like $m = 2^\alpha \cdot p_1^{l_1} \cdots p_k^{l_k}$, where $\alpha \in \{1,2,3,4,5,6,7,8\}$ and $m \leq 480$.

For (B), we have $|H| = \frac{n}{\gcd(n,m)} = \frac{2^5 \cdot 3 \cdot 5}{\gcd(n,m)}$, so we must have $m=2^\alpha \cdot p_1^{l_1} \cdots p_k^{l_k}$, where $\alpha \in \{0, 1,2,3,4\}$ and $m < 480$.

For (C), we have $|G: H| = \gcd(n,m)$, $m = 2^\alpha \cdot p_1^{l_1} \cdots p_k^{l_k}$, where $\alpha \in \{1,2,3,4,5,6,7,8\}$ and $p_1,p_2,\ldots,p_k \neq 3$, $m < 480$.

For (D), we have $|H| = \frac{n}{\gcd(n,m)} = \frac{2^5 \cdot 3 \cdot 5}{\gcd(n,m)}$. For $H$ to be divisible by $2$ and not by $3$. The prime factorization of $m$ should look like $2^\alpha \cdot 3^\beta \cdot p_1^{l_1} \cdots p_k^{l_k}$, where $\alpha \in \{0,1,2,3,4\}$ and $\beta \in \{1,2,3,4\}$, and $m < 480$.

mpavlov23
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  • Have you thought about which values of $m$ are actually possible? – Karl Mar 24 '24 at 14:21
  • I thought about that, and that's why I have given the factorization of m into primes because listing every number that is possible will be a lot of work, but I might be making a mistake somewhere, or could even make the range more precise. Please correct me if I am making a mistake. Thanks in advance! – mpavlov23 Mar 24 '24 at 16:56
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    Notice that $g^{\operatorname{gcd}(m,n)}=g^{am+bn}=(g^m)^ae^b\in H$, so $m=\operatorname{gcd}(m,n)$, i.e. $m|n$. The intuition that we must have $m|n$ comes from looking at the Cayley graph of $G$. – Karl Mar 24 '24 at 18:15
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    Thank you very much, that means that in the factorization of m, there can be only 2,3 and 5, which helps me narrow the possibilities for m in all cases, right? – mpavlov23 Mar 24 '24 at 20:17

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