# Is there any injective homomorphism (i.e. monomorphism) from a non-cyclic group of order $4$ to $\mathbb{Z}_8$?

The only such possible group is $$V$$ (up to isomorphism). If $$\phi$$ be such an into homomorphism, then $$o(\phi(V))=4$$ and $$\phi(V)$$ being a subgroup of $$\mathbb{Z}_8$$, it must be cyclic with a generator of order $$4$$. But, $$V$$ has not element of order $$4$$. A contradiction.

Does this work?

• What is an into homomorphism? – José Carlos Santos Apr 17 at 10:36
• A monomorphism. – Subhasis Biswas Apr 17 at 10:37
• i.e. $\forall y \in \phi(G)$, $\{x \in G: \phi(x)=y\}$ has only one element. – Subhasis Biswas Apr 17 at 10:40
• It is a nice exercise to prove that every subgroup of a cyclic group is cyclic. Your result follows immediately. – user1729 Apr 17 at 11:16

## 1 Answer

That argument works. Given a monomorphism $$f:G\to H$$ between groups $$G,H$$, then $$G$$ is isomorphic to $$f(G)$$, and the two must therefore have the same number of elements of any given order.

Alternatively, there is only one element of order $$2$$ in $$\Bbb Z_8$$, while there are three in $$V$$.

• Nice argument! Can you please post some links here to very similar group theory questions? (Contest math type). Thanks! – Subhasis Biswas Apr 17 at 11:01
• @SubhasisBiswas Any decent introductory group theory textbook should have lots of problems just like this one. – Arthur Apr 17 at 11:02
• I am going through. – Subhasis Biswas Apr 17 at 11:05