# $|G_1 |$ and $|G_2 |$ are coprime. Show that $K = H_1 \times H_2$

Let $$G_1$$ and $$G_2$$ be finite groups and let $$K \le G_1 \times G_2$$

(i) Set $$H_1 = \{ g \in G_1 : (g,e) \in K \}$$ and $$H_2 = \{ g \in G_2 : (e,g) \in K \}$$. Show that: $$H_1 \le G_1; \quad H_2 \le G_2; \quad H_1 \times H_2 \le K.$$

(ii) Suppose that $$|G_1|$$ and $$|G_2|$$ are coprime. Show that $$K = H_1 \times H_2$$.

(iii) Show that this result need not follow if $$|G_1|$$ and $$|G_2|$$ are not coprime.

I have done part (i). I was doing part (ii) and got stuck:

Since from above I showed that $$H_1 \times H_2 \subseteq K$$, now I only need to show that $$H_1 \times H_2 \supseteq K$$.

Let $$(g_1,g_2) \in K$$ where $$g_1 \in G_1$$ and $$g_2 \in G_2$$. Since $$K \leq G_1 \times G_2$$, we have $$(g_1,e) \in K$$ and $$(e,g_2) \in K$$. Hence $$g_1 \in H_1$$ and $$g_2 \in H_2$$.

Finally we have $$(g_1, g_2) \in H_1 \times H_2$$

As I haven't used the fact that order of $$G_1$$ and $$G_2$$ are coprime, there must be something fundamentally wrong with my proof.

• That $(g_1,g_2) \in K \Rightarrow (g_1,e) \in K$ was never explained in your "proof". Likewise with $(e,g_2) \in K$. Those are the only nontrivial parts of the argument.
– KCd
Apr 22, 2014 at 19:59
• (ii) Hint: What is the index of $H_1\times H_2$ in $K$ and how it compares with the indices of $H_1$ in $G_1$ and $H_2$ in $G_2$? Apr 22, 2014 at 20:33

Let $a = |G_1|$ be the order of $G_1$ and $b = |G_2|$ be the order of $G_2$. By Bézout's theorem, there are integers $u,v$ such that $au + bv = 1$. Let $(g_1, g_2) \in K$; we want to show that $g_1 \in H_1$ and $g_2 \in H_2$. But we have: $g_1^a = e \implies g_1^{au} = e = g_1^{1-bv} \implies g_1^{bv} = g_1$, and similarly $g_2^{au} = g_2$.
Thus $(g_1,e) = (g_1^{bv}, g_2^{bv}) = (g_1, g_2)^{bv}$ belongs to $K$ (because $(g_1,g_2)$ does), and so by definition $g_1 \in H_1$. Similarly using $g_2^{au} = g_2$, you get $g_2 \in H_2$.