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I am trying find the number of onto group homomorphisms from $U(20)$ onto $U(15)$ where $U(n):=\{1\leq r\leq n: (r, n)=1\}$.

Here is what I tried.

Since $U(20)\cong \mathbb Z_4\times \mathbb Z_2\cong U(15)$, it follows that only we need to find the number of onto group homomorphisms from $\mathbb Z_2\times \mathbb Z_4$ onto $\mathbb Z_2\times \mathbb Z_4$. Let $f$ be one such onto group homomorphism. Then by first isomorphism theorem we must have $$\frac{\mathbb Z_2\times \mathbb Z_4}{\ker f}\cong \mathbb Z_2\times \mathbb Z_4.$$

This tells that $|\ker f|=1$ and hence $f$ is an isomorphisms.

NOW ?

How many isomorphisms are there from $\mathbb Z_2\times \mathbb Z_4$ to $\mathbb Z_2\times \mathbb Z_4$: is it a valid question ? If so, what should I reply ?

Can someone show me what I am doing wrong please ?

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marked as duplicate by Gerry Myerson, N. F. Taussig, Joel Reyes Noche, hardmath, Community Sep 15 '15 at 17:50

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    $\begingroup$ Group isomorphisms must preserve order. Find the order 4 elements of each group, and work out the consequences of having order 4 elements of one group map to (distinct) order 4 elements of the other. $\endgroup$ – Gerry Myerson Sep 15 '15 at 7:01
  • $\begingroup$ There are four elements of order 4. hence number of isomorphisms is 4. Am I doing it correctly ? $\endgroup$ – Anjan3 Sep 15 '15 at 8:05
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    $\begingroup$ Does knowing where one element of order 4 goes, determine where everything else must go, Anjan? $\endgroup$ – Gerry Myerson Sep 15 '15 at 9:17