# In proving G contains an element of order 15 if contains normal subgroups of orders 3 and 5, respectively, is $HK$ itself cyclic with order 15?

In the proofs given by Brian Bi linked here and by Alec Mouri linked here, it is concluded that $$HK$$ is a subgroup and is isomorphic to $$H \times K$$ and then that $$HK$$ contains an element of order 15.

1. Is it correct that it is either:

• that the reason $$HK$$ contains an element of order 15 is that $$HK$$ is cyclic with order 15 or

• that that's not reason $$HK$$ contains an element of order 15, but $$HK$$ is still cyclic with order 15?

2. Actually, are groups isomorphic to cyclic groups also cyclic with the same order?

I know if images of cyclic groups under surjective homomorphisms are cyclic, but I didn't think they would be of different orders if we also assume injective.

The following are screenshots of the proofs mentioned above (Kiefer Sutherland's voice)  As soon as you've proved that $$HK\simeq H\times K$$, you know that $$H$$ is the product of two groups of prime order, hence $$H$$ and $$K$$ are abelian (cyclic) groups, and as their orders are coprime, you can apply the Chinese remainder theorem, which asserts the product is cyclic with order the product of the orders of each factor.
• Bernard, thank you! Is there some immediately obvious way to show $HK$ contains an element of order 15 other than saying that $H \times K$ contains an element of order 15? – user198044 Oct 14 '18 at 13:35
• Using the reverse isomorphism of the Chinese remainder theorem: of $a$ has order $3$ and $b$ has order $5$, the pair $(3,5)$ has order $15$ (in $H\times K$), and its image by the reverse isomorphism (in multiplicative notation) is $a^{-5}b^6$ (based on Bézout's identity: $\;2\cdot 3-5=1$), which can be simplified to $ab$. – Bernard Oct 14 '18 at 13:43
• My question is incorrect. Do you think the proofs conclude $HK$ contains an element of order 15 because $HK$ is cyclic with order 15? – user198044 Oct 14 '18 at 13:49