Find the validity of the following statement: There exists a monomorphism from a group of order $30$ into a group of order $100.$

Question

Find the validity of the following statement: There exists a monomorphism from a group of order $30$ into a group of order $100.$

I don't know how to solve these type of problems. Of course, this statement might be true but how to demonstrate a monomorphism between two arbitary groups, I have no idea about it. There should not exist any isomorphism between such groups but is a homomorphism at all possible? This seems absurd to me.

I found this problem in a handout which contained a number of problems,so, I have no idea about the source of this problem. There was no further details in it, so I myself cant provide further clarity to it. But still, this seemed interesting so I thought about posting it.

Answer 1

We know that a group $G$ of $30$ elements will have an subgroup $H$ of order $3$,say, $\lbrace 1,a,a^2 \rbrace$ (By Cauchy's theorem).

If there was a monomorphism into a group $G'$ of order $100$, what would be the order of the element $f(a)$ in $G'$? What about the order of the element $f(a^2)$ in $G'$?

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$\textbf{Edit:}$ What elements of a group are of order $1$?

Answer 2

If $f\colon G\to H$ is an homomorphism, then $f(G)\le H$ and hence (Lagrange) $|f(G)|$ divides $|H|$. If $f$ is also injective, then $|f(G)|=|G|$. But in your case $|G|=30$ does not divide $|H|=100$. So no, such a monomorphism doesn't exist.