Let $H$, $G_1$, $G_2$ be groups such that each $G_i$ $(i=1,2)$ has a subgroup $H_i$ isomorphic to $H$. Prove that the direct product $G_1 \times G_2$ has at least 3 distinct subgroups isomorphic to $H$.

I have zero idea how to approach this. I know that $H_1 \times H_2$ is subgroup isomorphic to $H$, but I don't know how to obtain the other two

EDIT: I now know that $H_1 \times H_2$ is wrong and that two groups that work are {$e_1$}x$H_2$, $H_1$x{e$_2$} are the other two, but I am unsure how to show that they are isomorphic to H since they are direct products. Is it enough to show f(uv)=f(u)f(v)

  • $\begingroup$ You need to assume that the group $H$ has more than one element; the statement is incorrect if $H$ is a group of order one. $\endgroup$ – bof Oct 30 '17 at 8:15
  • $\begingroup$ $[e_1}\times G_2$ and $G_1\times\{e_2\}$ are not isomorphic to $H,$ it's $\{e_1\}\times H_2$ and $H_1\times\{e_2\}$ that are isomorphic to $H.$ You also need a third group, and it's not $G_1\times G_2,$ $\endgroup$ – bof Oct 30 '17 at 8:19

The two you are looking for are just the subgroups $H_1 \times \{e_2\}$ and $\{e_1\} \times H_2$, where $e_1, e_2$ are the identity elements of $G_1$ and $G_2$. If $f$ and $g$ are the isomorphisms between $H_1 \leftrightarrow H$ and $H_2 \leftrightarrow H$, respectively, then the isomorphism between $(H_1 \times \{e_2\} )\leftrightarrow H$ is $(h_1,e_1) \mapsto f(h_1)$ and the isomorphism between $(\{e_1\} \times H_2 ) \leftrightarrow H$ is $(e,h_2) \mapsto g(h_2)$.

  • $\begingroup$ My bad, I wrote $H_1, H_2$ instead of $G_1, G_2$ in the problem. So in your answer should the $H_1, H_2$ be $G_1, G_2$? Also how do I show its an isomorphism? Is it just showing f(uv) = f(u)f(v)? $\endgroup$ – Vinny Chase Oct 30 '17 at 3:31
  • $\begingroup$ Sure, just swap out the group names. You'll have to show that it is a bijection and a homomorphism, so one part is showing that it is one-to-one and another is showing that f(uv) = f(u)f(v). $G_1 \times e_1$ is pretty simple, so don't overthink it -- it's basically a copy of $H$, just with an $e$ tacked on that doesn't do anything. $\endgroup$ – Andrew Tindall Oct 30 '17 at 3:57
  • $\begingroup$ Also, your example of $G_1 \times G_2$ is not quite correct. What you should use is the subgroup of $G_1 \times G_2$ that consists only of elements $(g,g)$; where the first element is identical to the second. $\endgroup$ – Andrew Tindall Oct 30 '17 at 3:58
  • $\begingroup$ Ok thank you! Does it matter that the group mentioned in your last comment could be empty? $\endgroup$ – Vinny Chase Oct 30 '17 at 4:01
  • $\begingroup$ Sorry: the elements themselves are not equal, their images under the isomorphisms are. I.e. they correspond to the same element in $H$. This is the subgroup of $G_1 \times G_2$ that consists only of elements $(f^{-1}h,g^{-1}h)$; where the first element is identical to the second under the two isomorphisms $f: G_1 \leftrightarrow H$ and $g: G_2 \leftrightarrow H$ $\endgroup$ – Andrew Tindall Oct 30 '17 at 4:03

$\def\zee{\mathbb Z}$ Let's do an example: $G_1 = H_1 = \zee$, $G_2 = H_2 = \zee$. and we use additive notation so the group identity is $0$. The three copies of $ \zee$ are $$\zee \times \{0\} = \{(n, 0) : n \in \zee\},$$ $$\{0\} \times \zee = \{(0, n) : n \in \zee\},$$ and the diagonal subgroup, $$D = \{(n, n) : n \in \zee\}.$$ Check that each of these is a subgroup isomorphic to $\zee$. If you understand this example, you can understand the general case. Important note: $\zee \times \zee$ is not isomorphic to $\zee$, and in the general case neither $G_1 \times G_2$ nor $H_1 \times H_2$ is isomorphic to $H$. In fact, if the groups are finite $H_1 \times H_2$ has order $|H_1 \times H_2| = |H|^2$, so the size is wrong.

  • $\begingroup$ I really don't understand this example. Could you explain it? I don't know how to construct the isomorphism between them. Also if the group $G_1$x$G_2$ isn't isomorphic to H how would I come up with something else that is? $\endgroup$ – Vinny Chase Oct 30 '17 at 3:59
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    $\begingroup$ Can you think of any reasonable map from the set of all integers to the set of all pairs of integers $(n, n)$ with both coordinates the same? $\endgroup$ – fredgoodman Oct 30 '17 at 16:29
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    $\begingroup$ $f: \mathbb{Z} ->\mathbb{Z} \times \mathbb{Z}$, $f(n) = (n,n)$? $\endgroup$ – Vinny Chase Oct 30 '17 at 16:51
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    $\begingroup$ Exactly. Now look at the definition of isomorphism and check whether $f$ is an isomorphism. $\endgroup$ – fredgoodman Oct 30 '17 at 17:01

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