Let $G_1, G_2, ... , G_t$ be finite groups, and define $G = G_1 × G_2 × \cdots × G_t$. If $G$ is cyclic then $G_1,G_2,...,G_t$ must also be cyclic. This is what I've done:
First, assume that $G$ is cyclic. Then there exists an element $g$ in $G$ such that $g$ generates $G$. Since $G$ is the direct product of the groups $G_1, G_2,..., G_t$, $g$ can be written as $(g_1, g_2, ..., g_t)$, where $g_i$ is an element in $G_i$ for each $i$. Since $g$ generates $G$, it follows that each $g_i$ generates $G_i$. Therefore, each $G_i$ is cyclic.
I'm not too sure if this is valid and formal proof. If this is right, how can I improve it to make it more formal and detailed? If this is not right, then what is wrong with my proof, and how do I prove this? I can't think of another way.
 A: Don't worry about the rude comments, @momo123321. Remember that everyone here is an almighty mathematical god who has been winning Fileds Medals sinc the age of 3, so and therefore is allowed to be rude. I will write an answer so that the rest of us, oh simple mortals, can get a bit of help from this site that they in all their wisdom and perfection have decided to give us, even if we are not deemed worthy of their help because we are not trying to solve something hard (a Millennium Problem, for example).
For some reason the code for curly braces ("\ {") is not working in the $\LaTeX$ environment, so I replaced curly braces ("{}") by square brackets ("[]").
We will use the fact that every subgroup of a cyclic group is itself cyclic, you can find a proof of this here. Therefore, in order to prove that $G_k$ is cyclic for $k\in [1,\ldots ,t]$ we only need to prove that it is isomorphic to a subgroup of $G$.
Fix $k\in [1,\ldots ,t]$, we will prove that $G_k$ is isomorphic to $[e_1]\times [e_2]\times \cdots \times [e_{k-1}]\times G_k\times [e_{k+1}]\times \cdots \times [e_t]$. Define $f:G_k\to [e_1]\times [e_2]\times \cdots \times [e_{k-1}]\times G_k\times [e_{k+1}]\times \cdots \times [e_t]$ as $f(g)=(e_1,e_2,\ldots ,e_{k-1},g,e_{k+1},\ldots ,e_t)$. By definition, every element of $[e_1]\times [e_2]\times \cdots \times [e_{k-1}]\times G_k\times [e_{k+1}]\times \cdots \times [e_t]$ has the form $(e_1,e_2,\ldots ,e_{k-1},g,e_{k+1},\ldots ,e_t)$, for some $g\in G_k$, since this last expression is $f(g)$, we conclude that $f$ is surjective. Now let $g_1,g_2\in G_k$ be such that $f(g_1)=f(g_2)$, then $(e_1,e_2,\ldots ,e_{k-1},g_1,e_{k+1},\ldots ,e_t)=(e_1,e_2,\ldots ,e_{k-1},g_2,e_{k+1},\ldots ,e_t)$, implying that $g_1=g_2$ (because it is an equality of $t$-tuples, so we must have equality componentwise), we conclude that $f$ is injective. Therefore, $f$ is bijective.
Finally, observe that if $g_1,g_2\in G_k$, then $f(g_1g_2)=(e_1,e_2,\ldots ,e_{k-1},g_1g_2,e_{k+1},\ldots ,e_t)$, but this is equal to $(e_1,e_2,\ldots ,e_{k-1},g_1,e_{k+1},\ldots ,e_t)(e_1,e_2,\ldots ,e_{k-1},g_2,e_{k+1},\ldots ,e_t)$, because the product of two elements in the direct product of groups is computed componentwise, this last expression is precisely $f(g_1)f(g_2)$. Therefore, $f$ is a group homomorphism.
We have a bijective group homomorphism $f:G_k\to [e_1]\times [e_2]\times \cdots \times [e_{k-1}]\times G_k\times [e_{k+1}]\times \cdots \times [e_t]$, therefore, we have an isomorphism, so we conclude that $G_k$ is isomorphic to $[e_1]\times [e_2]\times \cdots \times [e_{k-1}]\times G_k\times [e_{k+1}]\times \cdots \times [e_t]$. Therefore, $G_k$ is cyclic. Since we only used the fact that $k\in [1,\ldots ,t]$, this works for all $k\in [1,\ldots ,t]$, that is, $G_1,G_2,\ldots ,G_t$ are cyclic.
PS: I am not interested in upvotes or anything like that, but if this solution was useful for you or if there is something that you don't understand, please let me know it in the comments.
Also, consider joining to AoPS (Art of Problem Solving), the community is way nicer there. You will probably be interested in the College Math Forum.
