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We suppose that $G_1, G_2$ are finite groups with order $100$ and $f_1: G_1 \rightarrow \mathbb{Z}_{1200}, f_2: G_2 \rightarrow \mathbb{Z}_{120}^*$ are group homomorphisms. Which of the following statements are certainly wrong?

  • $|Ker(f_1)|=10$
  • $|Im(f_1)|=160$
  • $|Ker(f_2)|=30$
  • $|Im(f_2)|=20$

Could you give some hints what I am supposed to do?

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Hint:

If you have a group homomorphism $\phi:G_1 \rightarrow G_2$, then $Im(\phi) \subseteq G_2$ is a subgroup. As such, it must satisfy Lagrange's theorem.

Also, $\ker(\phi) \subseteq G_1$ is a (normal) subgroup.

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Well... by the First Group Isomorphism Theorem, $G/\ker(\phi) \cong \phi(G)$, whence $\frac{|G|}{|\ker(\phi)|} = |\phi(G)|$.

So, only the cases where $\ker(f_{1}) = 10$ and $Im(f_{2}) = 20$, since they are the two only cases that satisfy the divisibility criterion, are possible. However, as another answerer astutely points out (I didn't notice this at first), the second case does not work since the order of the range is $\phi(120) = 2^2\cdot 2\cdot 4 = 32$ and $20$ does not divide $32$.

Hence, the only possible case is where $\ker(f_{1}) = 10$.

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    $\begingroup$ $Z_{120}^\times$ has order $32$, not $120$. $\endgroup$ – Nishant Jun 7 '14 at 2:55
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A few different ideas are at play.

Firstly, both the kernel and the image of a group homorphism are groups (if you don't know this, then you should prove it!) and are subgroups. So the kernel of $f_1$ is a subgroup of $G_1$.

Secondly, you have Lagrange's Theorem, which gives you some limits on the sizes of subgroups if you know the size of the groups.

Thirdly, let's do a little experiment. If $G = {e}$ is the trivial group, and we have a homomorphism from $G$ to $\mathbb{Z}/2\mathbb{Z}$, the two-element cyclic group. Is it possible that the image is all of $\mathbb{Z}/2\mathbb{Z}$? Why or why not?

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