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I'm having some trouble with the following exercise:

Prove that, for all $m,n\in \mathbb N$: $$\gcd(m,n)\cdot \text{lcm}(m,n) = mn$$ To prove this use the the 2 proposition you proved before this exercice.

The two propositions where:

Let $G$ be a group, $H\leq G$ and $N \trianglelefteq G$. Then: $$H/(H\cap N) \simeq (HN)/N$$

Let $G$ be a group, $H \trianglelefteq G$ and $N \trianglelefteq G$, such that $N\subseteq H$. Then: $$(G/N)/(H/N)\simeq G/H$$

I was able to prove both propositions very easily using the first homomorphism theorem, but when it comes down to applying the proposition I always have a lot of struggle. How can this be solved?

Edit: I know that This post ask the same question but I'm asking for an alternative solution for this problem using these 2 specific propositions.

Eduardo Magalhães
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    Hint: You might find useful to think of $G=\mathbb{Z}$, and take as subgroups $m \mathbb{Z}$ and $n \mathbb{Z}$. The idea here is that $\mathbb{Z}/n\mathbb{Z}$ has cardinality $n$, and that $n\mathbb{Z} + m\mathbb{Z}=d\mathbb{Z}$ where $d$ is the gcd, $n \mathbb{Z} \cap m \mathbb{Z}=e\mathbb{Z}$ where $e$ is the lcm. – Nicolás Vilches Dec 13 '21 at 01:01
  • Please search first to avoid posting duplicate questions. – Bill Dubuque Dec 13 '21 at 01:11
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    How is this a duplicate question? The problem might be the same, in this question I'm exploring a different approach to solving said problem. Don't think this should be marked as a duplicate @BillDubuque – Eduardo Magalhães Dec 13 '21 at 01:12
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    Of course it is not a duplicate. @BillDubuque sees every arithmetic question as a duplicate of questions answered by him or his buddies. Staying on shoulders of giants does it to you. – markvs Dec 13 '21 at 01:26
  • @Eduardo For further details see e.g. here and here, which explicitly invoke (1) = 2nd isomorphism theorem. – Bill Dubuque Dec 13 '21 at 01:31
  • @markvs But it is a dupe and the OP has accepted one of the links I gave. – Bill Dubuque Dec 15 '21 at 08:46

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